On Statistics of Ordered Trees With a Prescribed Number of Allowed Children with at most 5 Children. By Lumpy (AKA Yonah's Computer) ------------------------------------------------------------------ On Statistics of Ordered Trees Where the Number of Children a Node Can Have is Restricted to {0, 1, 2, 3}. By Lumpy (AKA Yonah's Computer) This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 3}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^3; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^3+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 1/2*(208*n^3+816*n^2+608*n)/(13*n+25)/(n+3)/(2*n+7)*b[n]+1/2*(104*n^3+564*n^2+940*n+480)/(13*n+25)/(n+3)/(2*n+7)*b[n+1]+1/2*(143*n^3+990*n^2+2227*n+1620)/(13*n+25)/(n+3)/(2*n+7)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440, 81049178840528] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_1, the numbers of vertices with 0 and 1 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[3]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[3]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[3])+y[2]*f(x,y[0],y[1],y[2],y[3])^2+y[3]*f(x,y[0],y[1],y[2],y[3])^3). Our next step is to compute E[X_0] and E[X_1]. We do this by finding the total number of vertices with 0 and 1 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 1. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+3] = 1/2*(208*n^3+528*n^2+320*n)/(13*n+7)/(2*n+3)/(n+2)*b[n]+1/2*(104*n^3+316*n^2+280*n+68)/(13*n+7)/(2*n+3)/(n+2)*b[n+1]+1/2*(143*n^3+506*n^2+559*n+184)/(13*n+7)/(2*n+3)/(n+2)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 10, 31, 101, 336, 1128, 3823, 13051, 44803, 154518, 534964, 1858156, 6472168, 22597760, 79067375, 277164295, 973184313, 3422117190, 12049586631, 42478745781, 149915252028, 529606271560, 1872653175556, 6627147599476, 23471065878276, 83186110269928, 295024653043480, 1046972450508456] Doing the same for vertices with 1 children, we obtain the recurrence b[n+3] = 1/2*(208*n^3+608*n^2+400*n)/(2*n+5)/(13*n+12)/(n+1)*b[n]+1/2*(104*n^3+356*n^2+376*n+124)/(2*n+5)/(13*n+12)/(n+1)*b[n+1]+1/2*(143*n^3+561*n^2+676*n+240)/(2*n+5)/(13*n+12)/(n+1)*b[n+2] and the sequence [0, 1, 2, 6, 20, 65, 216, 728, 2472, 8451, 29050, 100298, 347568, 1208220, 4211312, 14712960, 51507280, 180642391, 634551606, 2232223626, 7862669700, 27727507521, 97884558992, 345891702456, 1223358393120, 4330360551700, 15339949125176, 54378725530392, 192893545292896, 684652948162760] Armed with this information, we can compute both the expectation E[X_0] = 12.917743 and E[X_1] = 8.447377. In order to calculate the variance of X_0 and X_1, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 2.411763, and making a similar calculation for 1 gives Var(X_1)= 6.107950. With these calculations finished, we begin demonstrating the fact that X_0 and X_1 are jointly normally distributed. In order to do so, we calculate E((X[0]-12.91774286)^p1*(X[1]-8.447376741)^p2/(2.411762577^(1/2*p1))/(6.107950496^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[1]*d/dy[1] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_1 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_1 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_1. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_1 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_1 is -0.927290, while the moment would be -0.927290 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_1 is -0.167368, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_1 is -2.751272, while the moment would be -2.781870 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_1 is -1.635317, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_1 is 0.138530, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_1 is 2.688331, while the moment would be 2.719734 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_1 is 1.492172, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_1 is 13.134336, while the moment would be 13.318404 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_1 is -2.746782, while the moment would be -2.781870 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_1 is -1.340854, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_1 is -12.894820, while the moment would be -13.129688 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_1 is -14.270563, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_1 is 1.173298, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_1 is 13.006971, while the moment would be 13.318404 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_1 is 13.219391, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_1 is 88.200403, while the moment would be 88.655330 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 3}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^3; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^3+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 1/2*(208*n^3+816*n^2+608*n)/(13*n+25)/(n+3)/(2*n+7)*b[n]+1/2*(104*n^3+564*n^2+940*n+480)/(13*n+25)/(n+3)/(2*n+7)*b[n+1]+1/2*(143*n^3+990*n^2+2227*n+1620)/(13*n+25)/(n+3)/(2*n+7)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440, 81049178840528] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_2, the numbers of vertices with 0 and 2 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[3]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[3]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[3])+y[2]*f(x,y[0],y[1],y[2],y[3])^2+y[3]*f(x,y[0],y[1],y[2],y[3])^3). Our next step is to compute E[X_0] and E[X_2]. We do this by finding the total number of vertices with 0 and 2 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 2. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+3] = 1/2*(208*n^3+528*n^2+320*n)/(13*n+7)/(2*n+3)/(n+2)*b[n]+1/2*(104*n^3+316*n^2+280*n+68)/(13*n+7)/(2*n+3)/(n+2)*b[n+1]+1/2*(143*n^3+506*n^2+559*n+184)/(13*n+7)/(2*n+3)/(n+2)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 10, 31, 101, 336, 1128, 3823, 13051, 44803, 154518, 534964, 1858156, 6472168, 22597760, 79067375, 277164295, 973184313, 3422117190, 12049586631, 42478745781, 149915252028, 529606271560, 1872653175556, 6627147599476, 23471065878276, 83186110269928, 295024653043480, 1046972450508456] Doing the same for vertices with 2 children, we obtain the recurrence b[n+3] = 1/2*(n+2)*(208*n^3+688*n^2+480*n)/n/(2*n+5)/(13*n+17)/(n+3)*b[n]+1/2*(n+2)*(104*n^3+396*n^2+472*n+180)/n/(2*n+5)/(13*n+17)/(n+3)*b[n+1]+1/2*(n+2)*(143*n^3+616*n^2+741*n+180)/n/(2*n+5)/(13*n+17)/(n+3)*b[n+2] and the sequence [0, 0, 1, 3, 10, 35, 120, 413, 1428, 4950, 17205, 59950, 209352, 732524, 2567656, 9014565, 31694120, 111578412, 393276555, 1387674063, 4901273250, 17327163735, 61307300516, 217087833028, 769259212560, 2727732391800, 9678382047100, 34360363904256, 122053568959728, 433776661421320] Armed with this information, we can compute both the expectation E[X_0] = 12.917743 and E[X_2] = 5.352018. In order to calculate the variance of X_0 and X_2, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 2.411763, and making a similar calculation for 2 gives Var(X_2)= 3.429405. With these calculations finished, we begin demonstrating the fact that X_0 and X_2 are jointly normally distributed. In order to do so, we calculate E((X[0]-12.91774286)^p1*(X[2]-5.352017967)^p2/(2.411762577^(1/2*p1))/(3.429404525^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[2]*d/dy[2] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_2 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_2 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_2. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_2 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_2 is -0.040768, while the moment would be -0.040768 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_2 is 0.037466, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_2 is -0.080352, while the moment would be -0.122303 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_2 is 0.211540, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_2 is -0.119086, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_2 is 0.967174, while the moment would be 1.003324 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_2 is -0.228557, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_2 is 2.609217, while the moment would be 3.019944 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_2 is -0.103199, while the moment would be -0.122303 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_2 is 0.039048, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_2 is -0.231977, while the moment would be -0.367315 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_2 is 0.431153, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_2 is -0.680877, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_2 is 2.827101, while the moment would be 3.019944 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_2 is -1.611396, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_2 is 7.433349, while the moment would be 9.119730 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 3}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^3; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^3+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 1/2*(208*n^3+816*n^2+608*n)/(13*n+25)/(n+3)/(2*n+7)*b[n]+1/2*(104*n^3+564*n^2+940*n+480)/(13*n+25)/(n+3)/(2*n+7)*b[n+1]+1/2*(143*n^3+990*n^2+2227*n+1620)/(13*n+25)/(n+3)/(2*n+7)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440, 81049178840528] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_3, the numbers of vertices with 0 and 3 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[3]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[3]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[3])+y[2]*f(x,y[0],y[1],y[2],y[3])^2+y[3]*f(x,y[0],y[1],y[2],y[3])^3). Our next step is to compute E[X_0] and E[X_3]. We do this by finding the total number of vertices with 0 and 3 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 3. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+3] = 1/2*(208*n^3+528*n^2+320*n)/(13*n+7)/(2*n+3)/(n+2)*b[n]+1/2*(104*n^3+316*n^2+280*n+68)/(13*n+7)/(2*n+3)/(n+2)*b[n+1]+1/2*(143*n^3+506*n^2+559*n+184)/(13*n+7)/(2*n+3)/(n+2)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 10, 31, 101, 336, 1128, 3823, 13051, 44803, 154518, 534964, 1858156, 6472168, 22597760, 79067375, 277164295, 973184313, 3422117190, 12049586631, 42478745781, 149915252028, 529606271560, 1872653175556, 6627147599476, 23471065878276, 83186110269928, 295024653043480, 1046972450508456] Doing the same for vertices with 3 children, we obtain the recurrence b[n+3] = 1/2*(n+2)*(208*n^3+768*n^2+560*n)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n]+1/2*(n+2)*(104*n^3+436*n^2+568*n+236)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n+1]+1/2*(n+2)*(143*n^3+671*n^2+754*n-56)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n+2] and the sequence [0, 0, 0, 1, 4, 15, 56, 203, 728, 2598, 9240, 32802, 116336, 412412, 1461824, 5181995, 18373616, 65166508, 231211152, 820654821, 2913977940, 10351141955, 36784590920, 130772586076, 465089770464, 1654708582200, 5889328479840, 20968345588320, 74681180866656, 266073305123304] Armed with this information, we can compute both the expectation E[X_0] = 12.917743 and E[X_3] = 3.282862. In order to calculate the variance of X_0 and X_3, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 2.411763, and making a similar calculation for 3 gives Var(X_3)= 1.518914. With these calculations finished, we begin demonstrating the fact that X_0 and X_3 are jointly normally distributed. In order to do so, we calculate E((X[0]-12.91774286)^p1*(X[3]-3.282862441)^p2/(2.411762577^(1/2*p1))/(1.518913920^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[3]*d/dy[3] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_3 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_3 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_3. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_3 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_3 is 0.660672, while the moment would be 0.660672 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_3 is 0.094336, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_3 is 1.897354, while the moment would be 1.982017 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_3 is 0.722482, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_3 is 0.026694, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_3 is 1.816689, while the moment would be 1.872976 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_3 is 0.504220, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_3 is 7.381147, while the moment would be 8.237857 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_3 is 1.939420, while the moment would be 1.982017 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_3 is 0.280034, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_3 is 6.994267, while the moment would be 7.676305 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_3 is 3.695857, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_3 is -0.102087, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_3 is 7.610422, while the moment would be 8.237857 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_3 is 2.346518, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_3 is 37.415862, while the moment would be 44.999663 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 3}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^3; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^3+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 1/2*(208*n^3+816*n^2+608*n)/(13*n+25)/(n+3)/(2*n+7)*b[n]+1/2*(104*n^3+564*n^2+940*n+480)/(13*n+25)/(n+3)/(2*n+7)*b[n+1]+1/2*(143*n^3+990*n^2+2227*n+1620)/(13*n+25)/(n+3)/(2*n+7)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440, 81049178840528] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_2, the numbers of vertices with 1 and 2 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[3]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[3]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[3])+y[2]*f(x,y[0],y[1],y[2],y[3])^2+y[3]*f(x,y[0],y[1],y[2],y[3])^3). Our next step is to compute E[X_1] and E[X_2]. We do this by finding the total number of vertices with 1 and 2 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 2. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+3] = 1/2*(208*n^3+608*n^2+400*n)/(2*n+5)/(13*n+12)/(n+1)*b[n]+1/2*(104*n^3+356*n^2+376*n+124)/(2*n+5)/(13*n+12)/(n+1)*b[n+1]+1/2*(143*n^3+561*n^2+676*n+240)/(2*n+5)/(13*n+12)/(n+1)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 6, 20, 65, 216, 728, 2472, 8451, 29050, 100298, 347568, 1208220, 4211312, 14712960, 51507280, 180642391, 634551606, 2232223626, 7862669700, 27727507521, 97884558992, 345891702456, 1223358393120, 4330360551700, 15339949125176, 54378725530392, 192893545292896, 684652948162760] Doing the same for vertices with 2 children, we obtain the recurrence b[n+3] = 1/2*(n+2)*(208*n^3+688*n^2+480*n)/n/(2*n+5)/(13*n+17)/(n+3)*b[n]+1/2*(n+2)*(104*n^3+396*n^2+472*n+180)/n/(2*n+5)/(13*n+17)/(n+3)*b[n+1]+1/2*(n+2)*(143*n^3+616*n^2+741*n+180)/n/(2*n+5)/(13*n+17)/(n+3)*b[n+2] and the sequence [0, 0, 1, 3, 10, 35, 120, 413, 1428, 4950, 17205, 59950, 209352, 732524, 2567656, 9014565, 31694120, 111578412, 393276555, 1387674063, 4901273250, 17327163735, 61307300516, 217087833028, 769259212560, 2727732391800, 9678382047100, 34360363904256, 122053568959728, 433776661421320] Armed with this information, we can compute both the expectation E[X_1] = 8.447377 and E[X_2] = 5.352018. In order to calculate the variance of X_1 and X_2, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.107950, and making a similar calculation for 2 gives Var(X_2)= 3.429405. With these calculations finished, we begin demonstrating the fact that X_1 and X_2 are jointly normally distributed. In order to do so, we calculate E((X[1]-8.447376741)^p1*(X[2]-5.352017967)^p2/(6.107950496^(1/2*p1))/(3.429404525^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[2]*d/dy[2] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_2 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_2 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_2. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_2 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_2 is -0.336229, while the moment would be -0.336229 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_2 is -0.084935, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_2 is -0.994141, while the moment would be -1.008687 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_2 is -0.659079, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_2 is -0.060748, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_2 is 1.203347, while the moment would be 1.226100 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_2 is 0.118571, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_2 is 4.052751, while the moment would be 4.356599 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_2 is -0.997647, while the moment would be -1.008687 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_2 is 0.047038, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_2 is -3.042166, while the moment would be -3.254125 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_2 is -1.039144, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_2 is -0.593352, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_2 is 4.121982, while the moment would be 4.356599 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_2 is -0.595973, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_2 is 15.073866, while the moment would be 17.446323 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 3}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^3; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^3+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 1/2*(208*n^3+816*n^2+608*n)/(13*n+25)/(n+3)/(2*n+7)*b[n]+1/2*(104*n^3+564*n^2+940*n+480)/(13*n+25)/(n+3)/(2*n+7)*b[n+1]+1/2*(143*n^3+990*n^2+2227*n+1620)/(13*n+25)/(n+3)/(2*n+7)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440, 81049178840528] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_3, the numbers of vertices with 1 and 3 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[3]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[3]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[3])+y[2]*f(x,y[0],y[1],y[2],y[3])^2+y[3]*f(x,y[0],y[1],y[2],y[3])^3). Our next step is to compute E[X_1] and E[X_3]. We do this by finding the total number of vertices with 1 and 3 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 3. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+3] = 1/2*(208*n^3+608*n^2+400*n)/(2*n+5)/(13*n+12)/(n+1)*b[n]+1/2*(104*n^3+356*n^2+376*n+124)/(2*n+5)/(13*n+12)/(n+1)*b[n+1]+1/2*(143*n^3+561*n^2+676*n+240)/(2*n+5)/(13*n+12)/(n+1)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 6, 20, 65, 216, 728, 2472, 8451, 29050, 100298, 347568, 1208220, 4211312, 14712960, 51507280, 180642391, 634551606, 2232223626, 7862669700, 27727507521, 97884558992, 345891702456, 1223358393120, 4330360551700, 15339949125176, 54378725530392, 192893545292896, 684652948162760] Doing the same for vertices with 3 children, we obtain the recurrence b[n+3] = 1/2*(n+2)*(208*n^3+768*n^2+560*n)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n]+1/2*(n+2)*(104*n^3+436*n^2+568*n+236)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n+1]+1/2*(n+2)*(143*n^3+671*n^2+754*n-56)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n+2] and the sequence [0, 0, 0, 1, 4, 15, 56, 203, 728, 2598, 9240, 32802, 116336, 412412, 1461824, 5181995, 18373616, 65166508, 231211152, 820654821, 2913977940, 10351141955, 36784590920, 130772586076, 465089770464, 1654708582200, 5889328479840, 20968345588320, 74681180866656, 266073305123304] Armed with this information, we can compute both the expectation E[X_1] = 8.447377 and E[X_3] = 3.282862. In order to calculate the variance of X_1 and X_3, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.107950, and making a similar calculation for 3 gives Var(X_3)= 1.518914. With these calculations finished, we begin demonstrating the fact that X_1 and X_3 are jointly normally distributed. In order to do so, we calculate E((X[1]-8.447376741)^p1*(X[3]-3.282862441)^p2/(6.107950496^(1/2*p1))/(1.518913920^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[3]*d/dy[3] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_3 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_3 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_3. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_3 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_3 is -0.331625, while the moment would be -0.331625 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_3 is -0.085927, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_3 is -0.968686, while the moment would be -0.994874 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_3 is -0.576424, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_3 is -0.059810, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_3 is 1.197169, while the moment would be 1.219950 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_3 is 0.056134, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_3 is 3.986841, while the moment would be 4.319700 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_3 is -0.983891, while the moment would be -0.994874 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_3 is 0.040619, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_3 is -2.970215, while the moment would be -3.203446 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_3 is -0.828099, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_3 is -0.584536, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_3 is 4.084104, while the moment would be 4.319700 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_3 is -0.740302, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_3 is 14.777079, while the moment would be 17.208469 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 3}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^3; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^3+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 1/2*(208*n^3+816*n^2+608*n)/(13*n+25)/(n+3)/(2*n+7)*b[n]+1/2*(104*n^3+564*n^2+940*n+480)/(13*n+25)/(n+3)/(2*n+7)*b[n+1]+1/2*(143*n^3+990*n^2+2227*n+1620)/(13*n+25)/(n+3)/(2*n+7)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440, 81049178840528] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_2 and X_3, the numbers of vertices with 2 and 3 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[3]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[3]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[3])+y[2]*f(x,y[0],y[1],y[2],y[3])^2+y[3]*f(x,y[0],y[1],y[2],y[3])^3). Our next step is to compute E[X_2] and E[X_3]. We do this by finding the total number of vertices with 2 and 3 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[2]*d/dy[2] to multiply every monomial by the number of nodes with 2 children in the corresponding tree, and similarly for 3. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 2 children over all trees with n vertices: b[n+3] = 1/2*(n+2)*(208*n^3+688*n^2+480*n)/n/(2*n+5)/(13*n+17)/(n+3)*b[n]+1/2*(n+2)*(104*n^3+396*n^2+472*n+180)/n/(2*n+5)/(13*n+17)/(n+3)*b[n+1]+1/2*(n+2)*(143*n^3+616*n^2+741*n+180)/n/(2*n+5)/(13*n+17)/(n+3)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 1, 3, 10, 35, 120, 413, 1428, 4950, 17205, 59950, 209352, 732524, 2567656, 9014565, 31694120, 111578412, 393276555, 1387674063, 4901273250, 17327163735, 61307300516, 217087833028, 769259212560, 2727732391800, 9678382047100, 34360363904256, 122053568959728, 433776661421320] Doing the same for vertices with 3 children, we obtain the recurrence b[n+3] = 1/2*(n+2)*(208*n^3+768*n^2+560*n)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n]+1/2*(n+2)*(104*n^3+436*n^2+568*n+236)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n+1]+1/2*(n+2)*(143*n^3+671*n^2+754*n-56)/(13*n+22)/(n-1)/(2*n+7)/(n+3)*b[n+2] and the sequence [0, 0, 0, 1, 4, 15, 56, 203, 728, 2598, 9240, 32802, 116336, 412412, 1461824, 5181995, 18373616, 65166508, 231211152, 820654821, 2913977940, 10351141955, 36784590920, 130772586076, 465089770464, 1654708582200, 5889328479840, 20968345588320, 74681180866656, 266073305123304] Armed with this information, we can compute both the expectation E[X_2] = 5.352018 and E[X_3] = 3.282862. In order to calculate the variance of X_2 and X_3, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[2]*d/dy[2] twice rather than once. Doing so gives Var(X_2)= 3.429405, and making a similar calculation for 3 gives Var(X_3)= 1.518914. With these calculations finished, we begin demonstrating the fact that X_2 and X_3 are jointly normally distributed. In order to do so, we calculate E((X[2]-5.352017967)^p1*(X[3]-3.282862441)^p2/(3.429404525^(1/2*p1))/(1.518913920^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[2]*d/dy[2] p1 times and y[3]*d/dy[3] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_2 and X_3 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_2 and X_3 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_2 and X_3. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_2 and X_3 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_2 and X_3 is -0.776984, while the moment would be -0.776984 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_2 and X_3 is -0.007983, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_2 and X_3 is -2.187851, while the moment would be -2.330953 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_2 and X_3 is -0.279083, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_2 and X_3 is -0.075900, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_2 and X_3 is 2.071856, while the moment would be 2.207409 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_2 and X_3 is -0.080376, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_2 and X_3 is 8.445305, while the moment would be 10.244455 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_2 and X_3 is -2.196059, while the moment would be -2.330953 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_2 and X_3 is 0.401573, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_2 and X_3 is -8.088825, while the moment would be -9.807273 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_2 and X_3 is 0.730806, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_2 and X_3 is -0.788546, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_2 and X_3 is 8.550379, while the moment would be 10.244455 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_2 and X_3 is -2.631729, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_2 and X_3 is 41.503153, while the moment would be 61.213755 if the random variables were normally distributed. On Statistics of Ordered Trees Where the Number of Children a Node Can Have is Restricted to {0, 1, 2, 4}. By Lumpy (AKA Yonah's Computer) This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(43433*n^5+400920*n^4+1311471*n^3+1779468*n^2+825484*n)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n]+1/3*(-29068*n^5-311922*n^4-1280196*n^3-2511614*n^2-2351192*n-836920)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+1]+1/3*(-3718*n^5-45474*n^4-218619*n^3-518413*n^2-608768*n-284228)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+2]+1/3*(18252*n^5+250614*n^4+1356006*n^3+3609351*n^2+4718739*n+2419002)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 10, 26, 72, 204, 593, 1753, 5263, 15995, 49127, 152231, 475359, 1494287, 4724903, 15017767, 47954400, 153764322, 494892393, 1598241869, 5177492708, 16820048006, 54785449703, 178873019471, 585312014446, 1919212203652, 6305054573239, 20750506952675] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_1, the numbers of vertices with 0 and 1 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[4])+y[2]*f(x,y[0],y[1],y[2],y[4])^2+y[4]*f(x,y[0],y[1],y[2],y[4])^4). Our next step is to compute E[X_0] and E[X_1]. We do this by finding the total number of vertices with 0 and 1 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 1. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+4] = 1/3*(130299*n^5+882024*n^4+2157001*n^3+2251320*n^2+846044*n)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n]+1/3*(-87204*n^5-633906*n^4-1738748*n^3-2214126*n^2-1276408*n-254328)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+1]+1/3*(-11154*n^5-86658*n^4-269237*n^3-425031*n^2-348148*n-120468)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+2]+1/3*(54756*n^5+452790*n^4+1425690*n^3+2097261*n^2+1390101*n+302802)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 7, 23, 71, 231, 743, 2423, 7927, 26113, 86373, 286859, 955683, 3192801, 10691957, 35880007, 120627751, 406213845, 1369933137, 4626130413, 15640674413, 52937700599, 179351745587, 608192667443, 2064144881171, 7010913942321, 23829761550013, 81050026263397, 275838652638453] Doing the same for vertices with 1 children, we obtain the recurrence b[n+4] = 1/3*(n+3)*(43433*n^5+314054*n^4+823685*n^3+924172*n^2+371108*n)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n]+1/3*(n+3)*(-29068*n^5-224718*n^4-656352*n^3-898254*n^2-569960*n-132408)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+1]+1/3*(n+3)*(-3718*n^5-30602*n^4-97069*n^3-148957*n^2-111976*n-34020)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+2]+1/3*(n+3)*(18252*n^5+159354*n^4+536070*n^3+862497*n^2+656859*n+185970)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+3] and the sequence [0, 1, 2, 6, 16, 50, 156, 504, 1632, 5337, 17530, 57893, 191940, 638651, 2131234, 7130385, 23908592, 80323351, 270319806, 911133600, 3075286440, 10392740253, 35161321118, 119082332284, 403681152144, 1369636242575, 4650698506246, 15803424390042, 53737941702256, 182846582623931] Armed with this information, we can compute both the expectation E[X_0] = 13.293104 and E[X_1] = 8.811668. In order to calculate the variance of X_0 and X_1, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.206628, and making a similar calculation for 1 gives Var(X_1)= 6.267966. With these calculations finished, we begin demonstrating the fact that X_0 and X_1 are jointly normally distributed. In order to do so, we calculate E((X[0]-13.29310427)^p1*(X[1]-8.811668218)^p2/(3.206627661^(1/2*p1))/(6.267965954^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[1]*d/dy[1] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_1 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_1 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_1. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_1 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_1 is -0.861541, while the moment would be -0.861541 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_1 is -0.145490, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_1 is -2.552802, while the moment would be -2.584623 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_1 is -1.420968, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_1 is 0.094296, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_1 is 2.451844, while the moment would be 2.484506 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_1 is 1.175769, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_1 is 11.645026, while the moment would be 11.907037 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_1 is -2.546427, while the moment would be -2.584623 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_1 is -0.933563, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_1 is -11.257906, while the moment would be -11.590759 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_1 is -10.609715, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_1 is 0.661009, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_1 is 11.464417, while the moment would be 11.907037 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_1 is 8.995288, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_1 is 72.840105, while the moment would be 75.664769 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(43433*n^5+400920*n^4+1311471*n^3+1779468*n^2+825484*n)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n]+1/3*(-29068*n^5-311922*n^4-1280196*n^3-2511614*n^2-2351192*n-836920)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+1]+1/3*(-3718*n^5-45474*n^4-218619*n^3-518413*n^2-608768*n-284228)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+2]+1/3*(18252*n^5+250614*n^4+1356006*n^3+3609351*n^2+4718739*n+2419002)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 10, 26, 72, 204, 593, 1753, 5263, 15995, 49127, 152231, 475359, 1494287, 4724903, 15017767, 47954400, 153764322, 494892393, 1598241869, 5177492708, 16820048006, 54785449703, 178873019471, 585312014446, 1919212203652, 6305054573239, 20750506952675] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_2, the numbers of vertices with 0 and 2 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[4])+y[2]*f(x,y[0],y[1],y[2],y[4])^2+y[4]*f(x,y[0],y[1],y[2],y[4])^4). Our next step is to compute E[X_0] and E[X_2]. We do this by finding the total number of vertices with 0 and 2 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 2. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+4] = 1/3*(130299*n^5+882024*n^4+2157001*n^3+2251320*n^2+846044*n)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n]+1/3*(-87204*n^5-633906*n^4-1738748*n^3-2214126*n^2-1276408*n-254328)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+1]+1/3*(-11154*n^5-86658*n^4-269237*n^3-425031*n^2-348148*n-120468)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+2]+1/3*(54756*n^5+452790*n^4+1425690*n^3+2097261*n^2+1390101*n+302802)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 7, 23, 71, 231, 743, 2423, 7927, 26113, 86373, 286859, 955683, 3192801, 10691957, 35880007, 120627751, 406213845, 1369933137, 4626130413, 15640674413, 52937700599, 179351745587, 608192667443, 2064144881171, 7010913942321, 23829761550013, 81050026263397, 275838652638453] Doing the same for vertices with 2 children, we obtain the recurrence b[n+4] = 1/3*(43433*n^5+334100*n^4+931625*n^3+1107670*n^2+466712*n)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n]+1/3*(-29068*n^5-238134*n^4-735300*n^3-1063350*n^2-713852*n-176736)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+1]+1/3*(-3718*n^5-32318*n^4-102643*n^3-139777*n^2-66010*n+4056)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+2]+1/3*(18252*n^5+167778*n^4+586110*n^3+960795*n^2+726123*n+196812)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+3] and the sequence [0, 0, 1, 3, 10, 30, 96, 308, 1008, 3312, 10950, 36322, 120912, 403624, 1350804, 4530540, 15224576, 51248064, 172770201, 583245147, 1971366090, 6670643406, 22594861905, 76604762619, 259940025612, 882740142900, 2999916948275, 10201858970121, 34715436142338, 118200887133910] Armed with this information, we can compute both the expectation E[X_0] = 13.293104 and E[X_2] = 5.696289. In order to calculate the variance of X_0 and X_2, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.206628, and making a similar calculation for 2 gives Var(X_2)= 3.754696. With these calculations finished, we begin demonstrating the fact that X_0 and X_2 are jointly normally distributed. In order to do so, we calculate E((X[0]-13.29310427)^p1*(X[2]-5.696289126)^p2/(3.206627661^(1/2*p1))/(3.754696002^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[2]*d/dy[2] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_2 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_2 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_2. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_2 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_2 is -0.178558, while the moment would be -0.178558 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_2 is 0.019744, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_2 is -0.465113, while the moment would be -0.535674 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_2 is 0.064954, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_2 is -0.123597, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_2 is 1.008825, while the moment would be 1.063766 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_2 is -0.270607, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_2 is 2.757497, while the moment would be 3.382596 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_2 is -0.498307, while the moment would be -0.535674 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_2 is 0.160583, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_2 is -1.271436, while the moment would be -1.641180 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_2 is 0.819323, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_2 is -0.702898, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_2 is 3.040237, while the moment would be 3.382596 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_2 is -1.918462, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_2 is 8.273553, while the moment would be 11.319970 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(43433*n^5+400920*n^4+1311471*n^3+1779468*n^2+825484*n)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n]+1/3*(-29068*n^5-311922*n^4-1280196*n^3-2511614*n^2-2351192*n-836920)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+1]+1/3*(-3718*n^5-45474*n^4-218619*n^3-518413*n^2-608768*n-284228)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+2]+1/3*(18252*n^5+250614*n^4+1356006*n^3+3609351*n^2+4718739*n+2419002)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 10, 26, 72, 204, 593, 1753, 5263, 15995, 49127, 152231, 475359, 1494287, 4724903, 15017767, 47954400, 153764322, 494892393, 1598241869, 5177492708, 16820048006, 54785449703, 178873019471, 585312014446, 1919212203652, 6305054573239, 20750506952675] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_4, the numbers of vertices with 0 and 4 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[4])+y[2]*f(x,y[0],y[1],y[2],y[4])^2+y[4]*f(x,y[0],y[1],y[2],y[4])^4). Our next step is to compute E[X_0] and E[X_4]. We do this by finding the total number of vertices with 0 and 4 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 4. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+4] = 1/3*(130299*n^5+882024*n^4+2157001*n^3+2251320*n^2+846044*n)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n]+1/3*(-87204*n^5-633906*n^4-1738748*n^3-2214126*n^2-1276408*n-254328)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+1]+1/3*(-11154*n^5-86658*n^4-269237*n^3-425031*n^2-348148*n-120468)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+2]+1/3*(54756*n^5+452790*n^4+1425690*n^3+2097261*n^2+1390101*n+302802)/(3*n+5)/(3*n+7)/(n+3)/(507*n^2+897*n+242)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 7, 23, 71, 231, 743, 2423, 7927, 26113, 86373, 286859, 955683, 3192801, 10691957, 35880007, 120627751, 406213845, 1369933137, 4626130413, 15640674413, 52937700599, 179351745587, 608192667443, 2064144881171, 7010913942321, 23829761550013, 81050026263397, 275838652638453] Doing the same for vertices with 4 children, we obtain the recurrence b[n+4] = 1/3*(43433*n^5+374192*n^4+1157271*n^3+1503964*n^2+677452*n)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n]+1/3*(-29068*n^5-264966*n^4-899732*n^3-1408194*n^2-999840*n-255480)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+1]+1/3*(-3718*n^5-35750*n^4-108543*n^3-60805*n^2+214192*n+256284)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+2]+1/3*(18252*n^5+184626*n^4+653790*n^3+862215*n^2+42783*n-500406)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+3] and the sequence [0, 0, 0, 0, 1, 5, 21, 77, 274, 954, 3300, 11352, 38940, 133276, 455546, 1555710, 5310176, 18120640, 61829748, 210974556, 719957310, 2457263046, 8388448662, 28642311654, 97822397376, 334177239600, 1141894993200, 3902896792080, 13343178515940, 45629086183956] Armed with this information, we can compute both the expectation E[X_0] = 13.293104 and E[X_4] = 2.198938. In order to calculate the variance of X_0 and X_4, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.206628, and making a similar calculation for 4 gives Var(X_4)= 0.911163. With these calculations finished, we begin demonstrating the fact that X_0 and X_4 are jointly normally distributed. In order to do so, we calculate E((X[0]-13.29310427)^p1*(X[4]-2.198938382)^p2/(3.206627661^(1/2*p1))/(.9111628099^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[4]*d/dy[4] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_4 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_4 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_4. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_4 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_4 is 0.746146, while the moment would be 0.746146 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_4 is 0.101120, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_4 is 2.124990, while the moment would be 2.238439 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_4 is 0.813207, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_4 is 0.063619, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_4 is 2.033126, while the moment would be 2.113469 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_4 is 0.703247, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_4 is 8.460375, while the moment would be 9.680811 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_4 is 2.175472, while the moment would be 2.238439 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_4 is 0.546059, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_4 is 8.223779, while the moment would be 9.207748 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_4 is 5.695669, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_4 is 0.280006, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_4 is 8.796556, while the moment would be 9.680811 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_4 is 4.735607, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_4 is 45.500002, while the moment would be 56.523742 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(43433*n^5+400920*n^4+1311471*n^3+1779468*n^2+825484*n)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n]+1/3*(-29068*n^5-311922*n^4-1280196*n^3-2511614*n^2-2351192*n-836920)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+1]+1/3*(-3718*n^5-45474*n^4-218619*n^3-518413*n^2-608768*n-284228)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+2]+1/3*(18252*n^5+250614*n^4+1356006*n^3+3609351*n^2+4718739*n+2419002)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 10, 26, 72, 204, 593, 1753, 5263, 15995, 49127, 152231, 475359, 1494287, 4724903, 15017767, 47954400, 153764322, 494892393, 1598241869, 5177492708, 16820048006, 54785449703, 178873019471, 585312014446, 1919212203652, 6305054573239, 20750506952675] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_2, the numbers of vertices with 1 and 2 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[4])+y[2]*f(x,y[0],y[1],y[2],y[4])^2+y[4]*f(x,y[0],y[1],y[2],y[4])^4). Our next step is to compute E[X_1] and E[X_2]. We do this by finding the total number of vertices with 1 and 2 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 2. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+4] = 1/3*(n+3)*(43433*n^5+314054*n^4+823685*n^3+924172*n^2+371108*n)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n]+1/3*(n+3)*(-29068*n^5-224718*n^4-656352*n^3-898254*n^2-569960*n-132408)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+1]+1/3*(n+3)*(-3718*n^5-30602*n^4-97069*n^3-148957*n^2-111976*n-34020)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+2]+1/3*(n+3)*(18252*n^5+159354*n^4+536070*n^3+862497*n^2+656859*n+185970)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 6, 16, 50, 156, 504, 1632, 5337, 17530, 57893, 191940, 638651, 2131234, 7130385, 23908592, 80323351, 270319806, 911133600, 3075286440, 10392740253, 35161321118, 119082332284, 403681152144, 1369636242575, 4650698506246, 15803424390042, 53737941702256, 182846582623931] Doing the same for vertices with 2 children, we obtain the recurrence b[n+4] = 1/3*(43433*n^5+334100*n^4+931625*n^3+1107670*n^2+466712*n)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n]+1/3*(-29068*n^5-238134*n^4-735300*n^3-1063350*n^2-713852*n-176736)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+1]+1/3*(-3718*n^5-32318*n^4-102643*n^3-139777*n^2-66010*n+4056)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+2]+1/3*(18252*n^5+167778*n^4+586110*n^3+960795*n^2+726123*n+196812)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+3] and the sequence [0, 0, 1, 3, 10, 30, 96, 308, 1008, 3312, 10950, 36322, 120912, 403624, 1350804, 4530540, 15224576, 51248064, 172770201, 583245147, 1971366090, 6670643406, 22594861905, 76604762619, 259940025612, 882740142900, 2999916948275, 10201858970121, 34715436142338, 118200887133910] Armed with this information, we can compute both the expectation E[X_1] = 8.811668 and E[X_2] = 5.696289. In order to calculate the variance of X_1 and X_2, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.267966, and making a similar calculation for 2 gives Var(X_2)= 3.754696. With these calculations finished, we begin demonstrating the fact that X_1 and X_2 are jointly normally distributed. In order to do so, we calculate E((X[1]-8.811668218)^p1*(X[2]-5.696289126)^p2/(6.267965954^(1/2*p1))/(3.754696002^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[2]*d/dy[2] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_2 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_2 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_2. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_2 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_2 is -0.345694, while the moment would be -0.345694 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_2 is -0.085926, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_2 is -1.023259, while the moment would be -1.037082 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_2 is -0.666284, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_2 is -0.058359, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_2 is 1.216629, while the moment would be 1.239009 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_2 is 0.129639, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_2 is 4.122981, while the moment would be 4.434051 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_2 is -1.024295, while the moment would be -1.037082 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_2 is 0.031305, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_2 is -3.146610, while the moment would be -3.359117 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_2 is -1.122788, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_2 is -0.570037, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_2 is 4.193724, while the moment would be 4.434051 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_2 is -0.505449, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_2 is 15.534449, while the moment would be 17.947059 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(43433*n^5+400920*n^4+1311471*n^3+1779468*n^2+825484*n)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n]+1/3*(-29068*n^5-311922*n^4-1280196*n^3-2511614*n^2-2351192*n-836920)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+1]+1/3*(-3718*n^5-45474*n^4-218619*n^3-518413*n^2-608768*n-284228)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+2]+1/3*(18252*n^5+250614*n^4+1356006*n^3+3609351*n^2+4718739*n+2419002)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 10, 26, 72, 204, 593, 1753, 5263, 15995, 49127, 152231, 475359, 1494287, 4724903, 15017767, 47954400, 153764322, 494892393, 1598241869, 5177492708, 16820048006, 54785449703, 178873019471, 585312014446, 1919212203652, 6305054573239, 20750506952675] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_4, the numbers of vertices with 1 and 4 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[4])+y[2]*f(x,y[0],y[1],y[2],y[4])^2+y[4]*f(x,y[0],y[1],y[2],y[4])^4). Our next step is to compute E[X_1] and E[X_4]. We do this by finding the total number of vertices with 1 and 4 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 4. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+4] = 1/3*(n+3)*(43433*n^5+314054*n^4+823685*n^3+924172*n^2+371108*n)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n]+1/3*(n+3)*(-29068*n^5-224718*n^4-656352*n^3-898254*n^2-569960*n-132408)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+1]+1/3*(n+3)*(-3718*n^5-30602*n^4-97069*n^3-148957*n^2-111976*n-34020)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+2]+1/3*(n+3)*(18252*n^5+159354*n^4+536070*n^3+862497*n^2+656859*n+185970)/(3*n+10)/(3*n+8)/(169*n^2+377*n+176)/(n+2)^2*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 6, 16, 50, 156, 504, 1632, 5337, 17530, 57893, 191940, 638651, 2131234, 7130385, 23908592, 80323351, 270319806, 911133600, 3075286440, 10392740253, 35161321118, 119082332284, 403681152144, 1369636242575, 4650698506246, 15803424390042, 53737941702256, 182846582623931] Doing the same for vertices with 4 children, we obtain the recurrence b[n+4] = 1/3*(43433*n^5+374192*n^4+1157271*n^3+1503964*n^2+677452*n)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n]+1/3*(-29068*n^5-264966*n^4-899732*n^3-1408194*n^2-999840*n-255480)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+1]+1/3*(-3718*n^5-35750*n^4-108543*n^3-60805*n^2+214192*n+256284)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+2]+1/3*(18252*n^5+184626*n^4+653790*n^3+862215*n^2+42783*n-500406)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+3] and the sequence [0, 0, 0, 0, 1, 5, 21, 77, 274, 954, 3300, 11352, 38940, 133276, 455546, 1555710, 5310176, 18120640, 61829748, 210974556, 719957310, 2457263046, 8388448662, 28642311654, 97822397376, 334177239600, 1141894993200, 3902896792080, 13343178515940, 45629086183956] Armed with this information, we can compute both the expectation E[X_1] = 8.811668 and E[X_4] = 2.198938. In order to calculate the variance of X_1 and X_4, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.267966, and making a similar calculation for 4 gives Var(X_4)= 0.911163. With these calculations finished, we begin demonstrating the fact that X_1 and X_4 are jointly normally distributed. In order to do so, we calculate E((X[1]-8.811668218)^p1*(X[4]-2.198938382)^p2/(6.267965954^(1/2*p1))/(.9111628099^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[4]*d/dy[4] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_4 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_4 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_4. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_4 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_4 is -0.304826, while the moment would be -0.304826 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_4 is -0.093599, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_4 is -0.889906, while the moment would be -0.914479 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_4 is -0.615967, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_4 is -0.051490, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_4 is 1.163919, while the moment would be 1.185838 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_4 is 0.082322, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_4 is 3.786446, while the moment would be 4.115029 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_4 is -0.903233, while the moment would be -0.914479 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_4 is -0.017406, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_4 is -2.695126, while the moment would be -2.913381 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_4 is -1.089786, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_4 is -0.502847, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_4 is 3.875884, while the moment would be 4.115029 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_4 is -0.441350, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_4 is 13.579454, while the moment would be 15.897386 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(43433*n^5+400920*n^4+1311471*n^3+1779468*n^2+825484*n)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n]+1/3*(-29068*n^5-311922*n^4-1280196*n^3-2511614*n^2-2351192*n-836920)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+1]+1/3*(-3718*n^5-45474*n^4-218619*n^3-518413*n^2-608768*n-284228)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+2]+1/3*(18252*n^5+250614*n^4+1356006*n^3+3609351*n^2+4718739*n+2419002)/(n+3)/(3*n+11)/(3*n+13)/(169*n^2+715*n+722)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 10, 26, 72, 204, 593, 1753, 5263, 15995, 49127, 152231, 475359, 1494287, 4724903, 15017767, 47954400, 153764322, 494892393, 1598241869, 5177492708, 16820048006, 54785449703, 178873019471, 585312014446, 1919212203652, 6305054573239, 20750506952675] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_2 and X_4, the numbers of vertices with 2 and 4 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[4])+y[2]*f(x,y[0],y[1],y[2],y[4])^2+y[4]*f(x,y[0],y[1],y[2],y[4])^4). Our next step is to compute E[X_2] and E[X_4]. We do this by finding the total number of vertices with 2 and 4 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[2]*d/dy[2] to multiply every monomial by the number of nodes with 2 children in the corresponding tree, and similarly for 4. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 2 children over all trees with n vertices: b[n+4] = 1/3*(43433*n^5+334100*n^4+931625*n^3+1107670*n^2+466712*n)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n]+1/3*(-29068*n^5-238134*n^4-735300*n^3-1063350*n^2-713852*n-176736)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+1]+1/3*(-3718*n^5-32318*n^4-102643*n^3-139777*n^2-66010*n+4056)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+2]+1/3*(18252*n^5+167778*n^4+586110*n^3+960795*n^2+726123*n+196812)/(3*n+11)/(3*n+7)/(n+1)/(169*n^2+455*n+284)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 1, 3, 10, 30, 96, 308, 1008, 3312, 10950, 36322, 120912, 403624, 1350804, 4530540, 15224576, 51248064, 172770201, 583245147, 1971366090, 6670643406, 22594861905, 76604762619, 259940025612, 882740142900, 2999916948275, 10201858970121, 34715436142338, 118200887133910] Doing the same for vertices with 4 children, we obtain the recurrence b[n+4] = 1/3*(43433*n^5+374192*n^4+1157271*n^3+1503964*n^2+677452*n)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n]+1/3*(-29068*n^5-264966*n^4-899732*n^3-1408194*n^2-999840*n-255480)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+1]+1/3*(-3718*n^5-35750*n^4-108543*n^3-60805*n^2+214192*n+256284)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+2]+1/3*(18252*n^5+184626*n^4+653790*n^3+862215*n^2+42783*n-500406)/(n-1)/(3*n+11)/(3*n+13)/(169*n^2+611*n+538)*b[n+3] and the sequence [0, 0, 0, 0, 1, 5, 21, 77, 274, 954, 3300, 11352, 38940, 133276, 455546, 1555710, 5310176, 18120640, 61829748, 210974556, 719957310, 2457263046, 8388448662, 28642311654, 97822397376, 334177239600, 1141894993200, 3902896792080, 13343178515940, 45629086183956] Armed with this information, we can compute both the expectation E[X_2] = 5.696289 and E[X_4] = 2.198938. In order to calculate the variance of X_2 and X_4, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[2]*d/dy[2] twice rather than once. Doing so gives Var(X_2)= 3.754696, and making a similar calculation for 4 gives Var(X_4)= 0.911163. With these calculations finished, we begin demonstrating the fact that X_2 and X_4 are jointly normally distributed. In order to do so, we calculate E((X[2]-5.696289126)^p1*(X[4]-2.198938382)^p2/(3.754696002^(1/2*p1))/(.9111628099^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[2]*d/dy[2] p1 times and y[4]*d/dy[4] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_2 and X_4 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_2 and X_4 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_2 and X_4. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_2 and X_4 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_2 and X_4 is -0.788313, while the moment would be -0.788313 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_2 and X_4 is -0.005499, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_2 and X_4 is -2.202877, while the moment would be -2.364939 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_2 and X_4 is -0.309250, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_2 and X_4 is -0.075645, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_2 and X_4 is 2.089698, while the moment would be 2.242874 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_2 and X_4 is -0.053011, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_2 and X_4 is 8.385083, while the moment would be 10.457247 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_2 and X_4 is -2.214440, while the moment would be -2.364939 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_2 and X_4 is 0.375487, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_2 and X_4 is -8.057990, while the moment would be -10.034138 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_2 and X_4 is 0.202603, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_2 and X_4 is -0.751912, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_2 and X_4 is 8.515923, while the moment would be 10.457247 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_2 and X_4 is -2.104303, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_2 and X_4 is 40.059844, while the moment would be 63.011901 if the random variables were normally distributed. On Statistics of Ordered Trees Where the Number of Children a Node Can Have is Restricted to {0, 1, 2, 5}. By Lumpy (AKA Yonah's Computer) This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(1244834736496*n^8+25173845861424*n^7+213189526142896*n^6+977188533380064*n^5+2607866752719904*n^4+4029727285844976*n^3+3310925556501504*n^2+1101137005014336*n)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n]+1/8*(400817410292*n^8+8706812671986*n^7+80839515009914*n^6+418113273345966*n^5+1314503735492420*n^4+2565573644768664*n^3+3026657439459486*n^2+1966334039565984*n+536326262980488)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+1]+1/8*(-686293858861*n^8-15937555839192*n^7-159700592295418*n^6-901597253172447*n^5-3135343431827266*n^4-6873979066695483*n^3-9272869387641207*n^2-7031596504053078*n-2292643196028648)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+2]+1/8*(-290622895680*n^8-7184965488480*n^7-77018316375180*n^6-467418597919470*n^5-1756013364152310*n^4-4179894777626310*n^3-6152753879142030*n^2-5116836201285420*n-1838909552825520)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+3]+1/8*(387497194240*n^8+10161199776000*n^7+115341549219760*n^6+739603068314400*n^5+2927162568472840*n^4+7312547814644940*n^3+11242304066709000*n^2+9704377428512940*n+3590736821453880)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 9, 22, 57, 155, 435, 1249, 3645, 10770, 32143, 96747, 293359, 895373, 2748803, 8483035, 26302248, 81896176, 255967640, 802790415, 2525691721, 7968972542, 25209580699, 79942927651, 254077293876, 809192984902, 2582113984084, 8254273128869] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_1, the numbers of vertices with 0 and 1 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[5])+y[2]*f(x,y[0],y[1],y[2],y[5])^2+y[5]*f(x,y[0],y[1],y[2],y[5])^5). Our next step is to compute E[X_0] and E[X_1]. We do this by finding the total number of vertices with 0 and 1 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 1. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+5] = 1/8*(4979338945984*n^8+76207379347216*n^7+488083146718384*n^6+1692064953156976*n^5+3420480088896016*n^4+4020360147414784*n^3+2536805421791616*n^2+661715516433024*n)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n]+1/8*(1603269641168*n^8+25339224863316*n^7+169124564583834*n^6+620772321972486*n^5+1366094335878438*n^4+1839770603822934*n^3+1477080817196416*n^2+645921051496464*n+117900214855344)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+1]+1/8*(-2745175435444*n^8-44759311860575*n^7-309764154588260*n^6-1182125397704690*n^5-2699900495706212*n^4-3734326321744475*n^3-2992998503184340*n^2-1219936508734260*n-173615486968944)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+2]+1/8*(-1162491582720*n^8-19535343447360*n^7-140055547743360*n^6-557502427535760*n^5-1340961306255240*n^4-1981489410352680*n^3-1737589168128600*n^2-812318732288280*n-150286387986000)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+3]+1/8*(1549988776960*n^8+26822118984960*n^7+196760801031360*n^6+794124789344400*n^5+1911000837338640*n^4+2767613012997480*n^3+2297266917461600*n^2+948037681721640*n+127435817442960)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 7, 19, 56, 171, 540, 1723, 5533, 17818, 57498, 185923, 602473, 1956566, 6367637, 20764907, 67837923, 221986035, 727468695, 2387109594, 7842294159, 25791611244, 84905797914, 279759215635, 922547710831, 3044552851031, 10054589313733, 33226892226862, 109870294035549] Doing the same for vertices with 1 children, we obtain the recurrence b[n+5] = 1/8*(n+4)*(1244834736496*n^8+20194506915440*n^7+137667988558624*n^6+510143316131888*n^5+1106938303606240*n^4+1402129662524096*n^3+956422883127312*n^2+269806524457248*n)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n]+1/8*(n+4)*(400817410292*n^8+6702725620526*n^7+47615533963138*n^6+186817615804400*n^5+440269292377738*n^4+633740411940212*n^3+539380144984100*n^2+245391772595838*n+44986737225708)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+1]+1/8*(n+4)*(-686293858861*n^8-11819792686026*n^7-86875751688068*n^6-354835952646233*n^5-877205961278883*n^4-1336018560633703*n^3-1212882196450626*n^2-590564317141296*n-114397565673600)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+2]+1/8*(n+4)*(-290622895680*n^8-5150605218720*n^7-39139735566540*n^6-166221618739890*n^5-430375971608850*n^4-693302731779570*n^3-675465847812450*n^2-361455480539100*n-80476789606560)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+3]+1/8*(n+4)*(387497194240*n^8+7061222222080*n^7+55063072226480*n^6+239239125414400*n^5+630753276634040*n^4+1027039388784940*n^3+999193314906340*n^2+522756932269920*n+109242991801440)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+4] and the sequence [0, 1, 2, 6, 16, 45, 132, 399, 1240, 3915, 12490, 40095, 129240, 417859, 1354458, 4400385, 14325968, 46729651, 152694630, 499742712, 1637923520, 5375320440, 17661389130, 58090909583, 191255341008, 630239517475, 2078516118926, 6860086934652, 22657403577256, 74881305538436] Armed with this information, we can compute both the expectation E[X_0] = 13.310717 and E[X_1] = 9.071823. In order to calculate the variance of X_0 and X_1, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.792702, and making a similar calculation for 1 gives Var(X_1)= 6.376825. With these calculations finished, we begin demonstrating the fact that X_0 and X_1 are jointly normally distributed. In order to do so, we calculate E((X[0]-13.31071704)^p1*(X[1]-9.071823087)^p2/(3.792701991^(1/2*p1))/(6.376824715^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[1]*d/dy[1] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_1 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_1 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_1. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_1 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_1 is -0.816487, while the moment would be -0.816487 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_1 is -0.131374, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_1 is -2.417276, while the moment would be -2.449462 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_1 is -1.282608, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_1 is 0.067071, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_1 is 2.300671, while the moment would be 2.333303 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_1 is 0.982171, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_1 is 10.702019, while the moment would be 10.999817 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_1 is -2.411149, while the moment would be -2.449462 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_1 is -0.698531, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_1 is -10.243885, while the moment would be -10.614260 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_1 is -8.500296, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_1 is 0.370480, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_1 is 10.510519, while the moment would be 10.999817 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_1 is 6.653139, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_1 is 63.893423, while the moment would be 67.665083 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(1244834736496*n^8+25173845861424*n^7+213189526142896*n^6+977188533380064*n^5+2607866752719904*n^4+4029727285844976*n^3+3310925556501504*n^2+1101137005014336*n)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n]+1/8*(400817410292*n^8+8706812671986*n^7+80839515009914*n^6+418113273345966*n^5+1314503735492420*n^4+2565573644768664*n^3+3026657439459486*n^2+1966334039565984*n+536326262980488)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+1]+1/8*(-686293858861*n^8-15937555839192*n^7-159700592295418*n^6-901597253172447*n^5-3135343431827266*n^4-6873979066695483*n^3-9272869387641207*n^2-7031596504053078*n-2292643196028648)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+2]+1/8*(-290622895680*n^8-7184965488480*n^7-77018316375180*n^6-467418597919470*n^5-1756013364152310*n^4-4179894777626310*n^3-6152753879142030*n^2-5116836201285420*n-1838909552825520)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+3]+1/8*(387497194240*n^8+10161199776000*n^7+115341549219760*n^6+739603068314400*n^5+2927162568472840*n^4+7312547814644940*n^3+11242304066709000*n^2+9704377428512940*n+3590736821453880)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 9, 22, 57, 155, 435, 1249, 3645, 10770, 32143, 96747, 293359, 895373, 2748803, 8483035, 26302248, 81896176, 255967640, 802790415, 2525691721, 7968972542, 25209580699, 79942927651, 254077293876, 809192984902, 2582113984084, 8254273128869] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_2, the numbers of vertices with 0 and 2 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[5])+y[2]*f(x,y[0],y[1],y[2],y[5])^2+y[5]*f(x,y[0],y[1],y[2],y[5])^5). Our next step is to compute E[X_0] and E[X_2]. We do this by finding the total number of vertices with 0 and 2 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 2. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+5] = 1/8*(4979338945984*n^8+76207379347216*n^7+488083146718384*n^6+1692064953156976*n^5+3420480088896016*n^4+4020360147414784*n^3+2536805421791616*n^2+661715516433024*n)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n]+1/8*(1603269641168*n^8+25339224863316*n^7+169124564583834*n^6+620772321972486*n^5+1366094335878438*n^4+1839770603822934*n^3+1477080817196416*n^2+645921051496464*n+117900214855344)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+1]+1/8*(-2745175435444*n^8-44759311860575*n^7-309764154588260*n^6-1182125397704690*n^5-2699900495706212*n^4-3734326321744475*n^3-2992998503184340*n^2-1219936508734260*n-173615486968944)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+2]+1/8*(-1162491582720*n^8-19535343447360*n^7-140055547743360*n^6-557502427535760*n^5-1340961306255240*n^4-1981489410352680*n^3-1737589168128600*n^2-812318732288280*n-150286387986000)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+3]+1/8*(1549988776960*n^8+26822118984960*n^7+196760801031360*n^6+794124789344400*n^5+1911000837338640*n^4+2767613012997480*n^3+2297266917461600*n^2+948037681721640*n+127435817442960)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 7, 19, 56, 171, 540, 1723, 5533, 17818, 57498, 185923, 602473, 1956566, 6367637, 20764907, 67837923, 221986035, 727468695, 2387109594, 7842294159, 25791611244, 84905797914, 279759215635, 922547710831, 3044552851031, 10054589313733, 33226892226862, 109870294035549] Doing the same for vertices with 2 children, we obtain the recurrence b[n+5] = 1/8*(n+4)*(4979338945984*n^8+85348675976304*n^7+616123387271536*n^6+2422540501839312*n^5+5586663564746128*n^4+7527663649885248*n^3+5459219058972096*n^2+1631432522234880*n)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n]+1/8*(n+4)*(1603269641168*n^8+28282580100892*n^7+212721427984526*n^6+885865402521910*n^5+2218238575357898*n^4+3389217883027414*n^3+3049161030775080*n^2+1453105112032296*n+274746673923840)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+1]+1/8*(n+4)*(-2745175435444*n^8-49799029627633*n^7-386415612040794*n^6-1671107751194888*n^5-4391163146593946*n^4-7146772723207947*n^3-6988358142574776*n^2-3713489122714092*n-805476863778480)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+2]+1/8*(n+4)*(-1162491582720*n^8-21669498302400*n^7-173294719687680*n^6-774586111963920*n^5-2110254365522520*n^4-3576285214379040*n^3-3668695149405960*n^2-2075809547326320*n-494929546272000)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+3]+1/8*(n+4)*(1549988776960*n^8+29667658791680*n^7+243472374854080*n^6+1116929523934640*n^5+3125987607029680*n^4+5450899317989480*n^3+5763256976406720*n^2+3362305370579640*n+823107483828000)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+4] and the sequence [0, 0, 1, 3, 10, 30, 90, 273, 840, 2628, 8325, 26620, 85668, 276926, 898079, 2919840, 9513000, 31051112, 101521467, 332425311, 1090002450, 3578531460, 11761905771, 38699236224, 127450069464, 420103593100, 1385865433875, 4575170995551, 15114420231498, 49963452087220] Armed with this information, we can compute both the expectation E[X_0] = 13.310717 and E[X_2] = 6.053041. In order to calculate the variance of X_0 and X_2, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.792702, and making a similar calculation for 2 gives Var(X_2)= 4.025753. With these calculations finished, we begin demonstrating the fact that X_0 and X_2 are jointly normally distributed. In order to do so, we calculate E((X[0]-13.31071704)^p1*(X[2]-6.053040808)^p2/(3.792701991^(1/2*p1))/(4.025752977^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[2]*d/dy[2] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_2 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_2 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_2. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_2 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_2 is -0.247560, while the moment would be -0.247560 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_2 is 0.013675, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_2 is -0.651367, while the moment would be -0.742679 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_2 is 0.032826, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_2 is -0.124591, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_2 is 1.051897, while the moment would be 1.122572 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_2 is -0.303150, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_2 is 2.920424, while the moment would be 3.735430 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_2 is -0.691508, while the moment would be -0.742679 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_2 is 0.246793, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_2 is -1.767233, while the moment would be -2.319070 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_2 is 1.246024, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_2 is -0.746881, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_2 is 3.246812, while the moment would be 3.735430 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_2 is -2.284262, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_2 is 8.874365, while the moment would be 13.502726 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(1244834736496*n^8+25173845861424*n^7+213189526142896*n^6+977188533380064*n^5+2607866752719904*n^4+4029727285844976*n^3+3310925556501504*n^2+1101137005014336*n)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n]+1/8*(400817410292*n^8+8706812671986*n^7+80839515009914*n^6+418113273345966*n^5+1314503735492420*n^4+2565573644768664*n^3+3026657439459486*n^2+1966334039565984*n+536326262980488)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+1]+1/8*(-686293858861*n^8-15937555839192*n^7-159700592295418*n^6-901597253172447*n^5-3135343431827266*n^4-6873979066695483*n^3-9272869387641207*n^2-7031596504053078*n-2292643196028648)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+2]+1/8*(-290622895680*n^8-7184965488480*n^7-77018316375180*n^6-467418597919470*n^5-1756013364152310*n^4-4179894777626310*n^3-6152753879142030*n^2-5116836201285420*n-1838909552825520)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+3]+1/8*(387497194240*n^8+10161199776000*n^7+115341549219760*n^6+739603068314400*n^5+2927162568472840*n^4+7312547814644940*n^3+11242304066709000*n^2+9704377428512940*n+3590736821453880)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 9, 22, 57, 155, 435, 1249, 3645, 10770, 32143, 96747, 293359, 895373, 2748803, 8483035, 26302248, 81896176, 255967640, 802790415, 2525691721, 7968972542, 25209580699, 79942927651, 254077293876, 809192984902, 2582113984084, 8254273128869] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_5, the numbers of vertices with 0 and 5 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[5])+y[2]*f(x,y[0],y[1],y[2],y[5])^2+y[5]*f(x,y[0],y[1],y[2],y[5])^5). Our next step is to compute E[X_0] and E[X_5]. We do this by finding the total number of vertices with 0 and 5 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 5. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+5] = 1/8*(4979338945984*n^8+76207379347216*n^7+488083146718384*n^6+1692064953156976*n^5+3420480088896016*n^4+4020360147414784*n^3+2536805421791616*n^2+661715516433024*n)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n]+1/8*(1603269641168*n^8+25339224863316*n^7+169124564583834*n^6+620772321972486*n^5+1366094335878438*n^4+1839770603822934*n^3+1477080817196416*n^2+645921051496464*n+117900214855344)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+1]+1/8*(-2745175435444*n^8-44759311860575*n^7-309764154588260*n^6-1182125397704690*n^5-2699900495706212*n^4-3734326321744475*n^3-2992998503184340*n^2-1219936508734260*n-173615486968944)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+2]+1/8*(-1162491582720*n^8-19535343447360*n^7-140055547743360*n^6-557502427535760*n^5-1340961306255240*n^4-1981489410352680*n^3-1737589168128600*n^2-812318732288280*n-150286387986000)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+3]+1/8*(1549988776960*n^8+26822118984960*n^7+196760801031360*n^6+794124789344400*n^5+1911000837338640*n^4+2767613012997480*n^3+2297266917461600*n^2+948037681721640*n+127435817442960)/(4*n+11)/(2*n+5)/(4*n+13)/(n+4)/(1210928732*n^4+6423635673*n^3+11236464238*n^2+7062753903*n+886723146)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 3, 7, 19, 56, 171, 540, 1723, 5533, 17818, 57498, 185923, 602473, 1956566, 6367637, 20764907, 67837923, 221986035, 727468695, 2387109594, 7842294159, 25791611244, 84905797914, 279759215635, 922547710831, 3044552851031, 10054589313733, 33226892226862, 109870294035549] Doing the same for vertices with 5 children, we obtain the recurrence b[n+5] = 1/8*(1244834736496*n^8+24765155229984*n^7+207413359877056*n^6+944788305027984*n^5+2516456700356704*n^4+3894433596605616*n^3+3212225074504944*n^2+1073352912611616*n)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n]+1/8*(400817410292*n^8+8174403239314*n^7+71257407389996*n^6+344359288233124*n^5+1000462686324062*n^4+1770431550273736*n^3+1838879323177254*n^2+1008695371051026*n+220660378495596)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+1]+1/8*(-686293858861*n^8-14339651569555*n^7-127548986663355*n^6-627006198262910*n^5-1845012386188568*n^4-3271346341617045*n^3-3297833186537970*n^2-1582899040284840*n-185885259330096)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+2]+1/8*(-290622895680*n^8-6217682646240*n^7-56224087336140*n^6-276429741085590*n^5-785699114152410*n^4-1239267451612170*n^3-845915724725850*n^2+152651419215480*n+376240288473000)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+3]+1/8*(387497194240*n^8+8483992125440*n^7+77876055285040*n^6+384789701860400*n^5+1081634436476560*n^4+1623077132697020*n^3+857502464402100*n^2-672067151983440*n-778708839933360)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+4] and the sequence [0, 0, 0, 0, 0, 1, 6, 28, 112, 414, 1462, 5027, 17028, 57200, 191282, 638106, 2125776, 7075944, 23540580, 78286802, 260284876, 865243071, 2876003438, 9559397287, 31774891368, 105625297520, 351152530820, 1167556333320, 3882589502820, 12913142204865] Armed with this information, we can compute both the expectation E[X_0] = 13.310717 and E[X_5] = 1.564419. In order to calculate the variance of X_0 and X_5, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.792702, and making a similar calculation for 5 gives Var(X_5)= 0.609571. With these calculations finished, we begin demonstrating the fact that X_0 and X_5 are jointly normally distributed. In order to do so, we calculate E((X[0]-13.31071704)^p1*(X[5]-1.564419059)^p2/(3.792701991^(1/2*p1))/(.6095706484^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[5]*d/dy[5] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_5 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_5 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_5. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_5 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_5 is 0.782644, while the moment would be 0.782644 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_5 is 0.108370, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_5 is 2.213483, while the moment would be 2.347932 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_5 is 0.920923, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_5 is 0.084686, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_5 is 2.127135, while the moment would be 2.225063 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_5 is 0.849495, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_5 is 8.857199, while the moment would be 10.350377 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_5 is 2.270551, while the moment would be 2.347932 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_5 is 0.730398, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_5 is 8.708759, while the moment would be 9.920159 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_5 is 7.216770, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_5 is 0.528064, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_5 is 9.276801, while the moment would be 10.350377 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_5 is 6.460333, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_5 is 48.291106, while the moment would be 62.106938 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(1244834736496*n^8+25173845861424*n^7+213189526142896*n^6+977188533380064*n^5+2607866752719904*n^4+4029727285844976*n^3+3310925556501504*n^2+1101137005014336*n)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n]+1/8*(400817410292*n^8+8706812671986*n^7+80839515009914*n^6+418113273345966*n^5+1314503735492420*n^4+2565573644768664*n^3+3026657439459486*n^2+1966334039565984*n+536326262980488)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+1]+1/8*(-686293858861*n^8-15937555839192*n^7-159700592295418*n^6-901597253172447*n^5-3135343431827266*n^4-6873979066695483*n^3-9272869387641207*n^2-7031596504053078*n-2292643196028648)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+2]+1/8*(-290622895680*n^8-7184965488480*n^7-77018316375180*n^6-467418597919470*n^5-1756013364152310*n^4-4179894777626310*n^3-6152753879142030*n^2-5116836201285420*n-1838909552825520)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+3]+1/8*(387497194240*n^8+10161199776000*n^7+115341549219760*n^6+739603068314400*n^5+2927162568472840*n^4+7312547814644940*n^3+11242304066709000*n^2+9704377428512940*n+3590736821453880)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 9, 22, 57, 155, 435, 1249, 3645, 10770, 32143, 96747, 293359, 895373, 2748803, 8483035, 26302248, 81896176, 255967640, 802790415, 2525691721, 7968972542, 25209580699, 79942927651, 254077293876, 809192984902, 2582113984084, 8254273128869] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_2, the numbers of vertices with 1 and 2 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[5])+y[2]*f(x,y[0],y[1],y[2],y[5])^2+y[5]*f(x,y[0],y[1],y[2],y[5])^5). Our next step is to compute E[X_1] and E[X_2]. We do this by finding the total number of vertices with 1 and 2 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 2. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+5] = 1/8*(n+4)*(1244834736496*n^8+20194506915440*n^7+137667988558624*n^6+510143316131888*n^5+1106938303606240*n^4+1402129662524096*n^3+956422883127312*n^2+269806524457248*n)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n]+1/8*(n+4)*(400817410292*n^8+6702725620526*n^7+47615533963138*n^6+186817615804400*n^5+440269292377738*n^4+633740411940212*n^3+539380144984100*n^2+245391772595838*n+44986737225708)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+1]+1/8*(n+4)*(-686293858861*n^8-11819792686026*n^7-86875751688068*n^6-354835952646233*n^5-877205961278883*n^4-1336018560633703*n^3-1212882196450626*n^2-590564317141296*n-114397565673600)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+2]+1/8*(n+4)*(-290622895680*n^8-5150605218720*n^7-39139735566540*n^6-166221618739890*n^5-430375971608850*n^4-693302731779570*n^3-675465847812450*n^2-361455480539100*n-80476789606560)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+3]+1/8*(n+4)*(387497194240*n^8+7061222222080*n^7+55063072226480*n^6+239239125414400*n^5+630753276634040*n^4+1027039388784940*n^3+999193314906340*n^2+522756932269920*n+109242991801440)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 6, 16, 45, 132, 399, 1240, 3915, 12490, 40095, 129240, 417859, 1354458, 4400385, 14325968, 46729651, 152694630, 499742712, 1637923520, 5375320440, 17661389130, 58090909583, 191255341008, 630239517475, 2078516118926, 6860086934652, 22657403577256, 74881305538436] Doing the same for vertices with 2 children, we obtain the recurrence b[n+5] = 1/8*(n+4)*(4979338945984*n^8+85348675976304*n^7+616123387271536*n^6+2422540501839312*n^5+5586663564746128*n^4+7527663649885248*n^3+5459219058972096*n^2+1631432522234880*n)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n]+1/8*(n+4)*(1603269641168*n^8+28282580100892*n^7+212721427984526*n^6+885865402521910*n^5+2218238575357898*n^4+3389217883027414*n^3+3049161030775080*n^2+1453105112032296*n+274746673923840)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+1]+1/8*(n+4)*(-2745175435444*n^8-49799029627633*n^7-386415612040794*n^6-1671107751194888*n^5-4391163146593946*n^4-7146772723207947*n^3-6988358142574776*n^2-3713489122714092*n-805476863778480)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+2]+1/8*(n+4)*(-1162491582720*n^8-21669498302400*n^7-173294719687680*n^6-774586111963920*n^5-2110254365522520*n^4-3576285214379040*n^3-3668695149405960*n^2-2075809547326320*n-494929546272000)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+3]+1/8*(n+4)*(1549988776960*n^8+29667658791680*n^7+243472374854080*n^6+1116929523934640*n^5+3125987607029680*n^4+5450899317989480*n^3+5763256976406720*n^2+3362305370579640*n+823107483828000)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+4] and the sequence [0, 0, 1, 3, 10, 30, 90, 273, 840, 2628, 8325, 26620, 85668, 276926, 898079, 2919840, 9513000, 31051112, 101521467, 332425311, 1090002450, 3578531460, 11761905771, 38699236224, 127450069464, 420103593100, 1385865433875, 4575170995551, 15114420231498, 49963452087220] Armed with this information, we can compute both the expectation E[X_1] = 9.071823 and E[X_2] = 6.053041. In order to calculate the variance of X_1 and X_2, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.376825, and making a similar calculation for 2 gives Var(X_2)= 4.025753. With these calculations finished, we begin demonstrating the fact that X_1 and X_2 are jointly normally distributed. In order to do so, we calculate E((X[1]-9.071823087)^p1*(X[2]-6.053040808)^p2/(6.376824715^(1/2*p1))/(4.025752977^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[2]*d/dy[2] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_2 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_2 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_2. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_2 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_2 is -0.357262, while the moment would be -0.357262 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_2 is -0.084978, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_2 is -1.057038, while the moment would be -1.071786 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_2 is -0.653788, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_2 is -0.057289, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_2 is 1.233261, while the moment would be 1.255272 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_2 is 0.126732, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_2 is 4.205823, while the moment would be 4.531635 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_2 is -1.057493, while the moment would be -1.071786 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_2 is 0.027923, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_2 is -3.272537, while the moment would be -3.488957 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_2 is -1.114707, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_2 is -0.559829, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_2 is 4.287624, while the moment would be 4.531635 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_2 is -0.496322, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_2 is 16.108718, while the moment would be 18.580793 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(1244834736496*n^8+25173845861424*n^7+213189526142896*n^6+977188533380064*n^5+2607866752719904*n^4+4029727285844976*n^3+3310925556501504*n^2+1101137005014336*n)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n]+1/8*(400817410292*n^8+8706812671986*n^7+80839515009914*n^6+418113273345966*n^5+1314503735492420*n^4+2565573644768664*n^3+3026657439459486*n^2+1966334039565984*n+536326262980488)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+1]+1/8*(-686293858861*n^8-15937555839192*n^7-159700592295418*n^6-901597253172447*n^5-3135343431827266*n^4-6873979066695483*n^3-9272869387641207*n^2-7031596504053078*n-2292643196028648)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+2]+1/8*(-290622895680*n^8-7184965488480*n^7-77018316375180*n^6-467418597919470*n^5-1756013364152310*n^4-4179894777626310*n^3-6152753879142030*n^2-5116836201285420*n-1838909552825520)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+3]+1/8*(387497194240*n^8+10161199776000*n^7+115341549219760*n^6+739603068314400*n^5+2927162568472840*n^4+7312547814644940*n^3+11242304066709000*n^2+9704377428512940*n+3590736821453880)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 9, 22, 57, 155, 435, 1249, 3645, 10770, 32143, 96747, 293359, 895373, 2748803, 8483035, 26302248, 81896176, 255967640, 802790415, 2525691721, 7968972542, 25209580699, 79942927651, 254077293876, 809192984902, 2582113984084, 8254273128869] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_5, the numbers of vertices with 1 and 5 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[5])+y[2]*f(x,y[0],y[1],y[2],y[5])^2+y[5]*f(x,y[0],y[1],y[2],y[5])^5). Our next step is to compute E[X_1] and E[X_5]. We do this by finding the total number of vertices with 1 and 5 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 5. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+5] = 1/8*(n+4)*(1244834736496*n^8+20194506915440*n^7+137667988558624*n^6+510143316131888*n^5+1106938303606240*n^4+1402129662524096*n^3+956422883127312*n^2+269806524457248*n)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n]+1/8*(n+4)*(400817410292*n^8+6702725620526*n^7+47615533963138*n^6+186817615804400*n^5+440269292377738*n^4+633740411940212*n^3+539380144984100*n^2+245391772595838*n+44986737225708)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+1]+1/8*(n+4)*(-686293858861*n^8-11819792686026*n^7-86875751688068*n^6-354835952646233*n^5-877205961278883*n^4-1336018560633703*n^3-1212882196450626*n^2-590564317141296*n-114397565673600)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+2]+1/8*(n+4)*(-290622895680*n^8-5150605218720*n^7-39139735566540*n^6-166221618739890*n^5-430375971608850*n^4-693302731779570*n^3-675465847812450*n^2-361455480539100*n-80476789606560)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+3]+1/8*(n+4)*(387497194240*n^8+7061222222080*n^7+55063072226480*n^6+239239125414400*n^5+630753276634040*n^4+1027039388784940*n^3+999193314906340*n^2+522756932269920*n+109242991801440)/(4*n+11)/(2*n+7)/(4*n+17)/(302732183*n^4+1883793665*n^3+4113406714*n^2+3627543037*n+1008262260)/(n+3)^2*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 6, 16, 45, 132, 399, 1240, 3915, 12490, 40095, 129240, 417859, 1354458, 4400385, 14325968, 46729651, 152694630, 499742712, 1637923520, 5375320440, 17661389130, 58090909583, 191255341008, 630239517475, 2078516118926, 6860086934652, 22657403577256, 74881305538436] Doing the same for vertices with 5 children, we obtain the recurrence b[n+5] = 1/8*(1244834736496*n^8+24765155229984*n^7+207413359877056*n^6+944788305027984*n^5+2516456700356704*n^4+3894433596605616*n^3+3212225074504944*n^2+1073352912611616*n)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n]+1/8*(400817410292*n^8+8174403239314*n^7+71257407389996*n^6+344359288233124*n^5+1000462686324062*n^4+1770431550273736*n^3+1838879323177254*n^2+1008695371051026*n+220660378495596)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+1]+1/8*(-686293858861*n^8-14339651569555*n^7-127548986663355*n^6-627006198262910*n^5-1845012386188568*n^4-3271346341617045*n^3-3297833186537970*n^2-1582899040284840*n-185885259330096)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+2]+1/8*(-290622895680*n^8-6217682646240*n^7-56224087336140*n^6-276429741085590*n^5-785699114152410*n^4-1239267451612170*n^3-845915724725850*n^2+152651419215480*n+376240288473000)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+3]+1/8*(387497194240*n^8+8483992125440*n^7+77876055285040*n^6+384789701860400*n^5+1081634436476560*n^4+1623077132697020*n^3+857502464402100*n^2-672067151983440*n-778708839933360)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+4] and the sequence [0, 0, 0, 0, 0, 1, 6, 28, 112, 414, 1462, 5027, 17028, 57200, 191282, 638106, 2125776, 7075944, 23540580, 78286802, 260284876, 865243071, 2876003438, 9559397287, 31774891368, 105625297520, 351152530820, 1167556333320, 3882589502820, 12913142204865] Armed with this information, we can compute both the expectation E[X_1] = 9.071823 and E[X_5] = 1.564419. In order to calculate the variance of X_1 and X_5, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.376825, and making a similar calculation for 5 gives Var(X_5)= 0.609571. With these calculations finished, we begin demonstrating the fact that X_1 and X_5 are jointly normally distributed. In order to do so, we calculate E((X[1]-9.071823087)^p1*(X[5]-1.564419059)^p2/(6.376824715^(1/2*p1))/(.6095706484^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[5]*d/dy[5] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_5 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_5 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_5. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_5 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_5 is -0.279628, while the moment would be -0.279628 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_5 is -0.098703, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_5 is -0.817505, while the moment would be -0.838883 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_5 is -0.647907, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_5 is -0.045118, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_5 is 1.135586, while the moment would be 1.156383 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_5 is 0.102515, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_5 is 3.612180, while the moment would be 3.938299 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_5 is -0.827995, while the moment would be -0.838883 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_5 is -0.058032, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_5 is -2.448911, while the moment would be -2.647835 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_5 is -1.272276, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_5 is -0.440155, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_5 is 3.702137, while the moment would be 3.938299 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_5 is -0.230581, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_5 is 12.592174, while the moment would be 14.776526 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 2, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^2+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^2+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(1244834736496*n^8+25173845861424*n^7+213189526142896*n^6+977188533380064*n^5+2607866752719904*n^4+4029727285844976*n^3+3310925556501504*n^2+1101137005014336*n)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n]+1/8*(400817410292*n^8+8706812671986*n^7+80839515009914*n^6+418113273345966*n^5+1314503735492420*n^4+2565573644768664*n^3+3026657439459486*n^2+1966334039565984*n+536326262980488)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+1]+1/8*(-686293858861*n^8-15937555839192*n^7-159700592295418*n^6-901597253172447*n^5-3135343431827266*n^4-6873979066695483*n^3-9272869387641207*n^2-7031596504053078*n-2292643196028648)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+2]+1/8*(-290622895680*n^8-7184965488480*n^7-77018316375180*n^6-467418597919470*n^5-1756013364152310*n^4-4179894777626310*n^3-6152753879142030*n^2-5116836201285420*n-1838909552825520)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+3]+1/8*(387497194240*n^8+10161199776000*n^7+115341549219760*n^6+739603068314400*n^5+2927162568472840*n^4+7312547814644940*n^3+11242304066709000*n^2+9704377428512940*n+3590736821453880)/(2*n+9)/(4*n+21)/(n+4)/(4*n+15)/(302732183*n^4+3094722397*n^3+11581180807*n^2+18716666192*n+10935737859)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 2, 4, 9, 22, 57, 155, 435, 1249, 3645, 10770, 32143, 96747, 293359, 895373, 2748803, 8483035, 26302248, 81896176, 255967640, 802790415, 2525691721, 7968972542, 25209580699, 79942927651, 254077293876, 809192984902, 2582113984084, 8254273128869] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_2 and X_5, the numbers of vertices with 2 and 5 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[2], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[2],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[2],y[5])+y[2]*f(x,y[0],y[1],y[2],y[5])^2+y[5]*f(x,y[0],y[1],y[2],y[5])^5). Our next step is to compute E[X_2] and E[X_5]. We do this by finding the total number of vertices with 2 and 5 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[2]*d/dy[2] to multiply every monomial by the number of nodes with 2 children in the corresponding tree, and similarly for 5. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 2 children over all trees with n vertices: b[n+5] = 1/8*(n+4)*(4979338945984*n^8+85348675976304*n^7+616123387271536*n^6+2422540501839312*n^5+5586663564746128*n^4+7527663649885248*n^3+5459219058972096*n^2+1631432522234880*n)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n]+1/8*(n+4)*(1603269641168*n^8+28282580100892*n^7+212721427984526*n^6+885865402521910*n^5+2218238575357898*n^4+3389217883027414*n^3+3049161030775080*n^2+1453105112032296*n+274746673923840)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+1]+1/8*(n+4)*(-2745175435444*n^8-49799029627633*n^7-386415612040794*n^6-1671107751194888*n^5-4391163146593946*n^4-7146772723207947*n^3-6988358142574776*n^2-3713489122714092*n-805476863778480)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+2]+1/8*(n+4)*(-1162491582720*n^8-21669498302400*n^7-173294719687680*n^6-774586111963920*n^5-2110254365522520*n^4-3576285214379040*n^3-3668695149405960*n^2-2075809547326320*n-494929546272000)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+3]+1/8*(n+4)*(1549988776960*n^8+29667658791680*n^7+243472374854080*n^6+1116929523934640*n^5+3125987607029680*n^4+5450899317989480*n^3+5763256976406720*n^2+3362305370579640*n+823107483828000)/(2*n+9)/(4*n+13)/(n+3)/(n+2)/(4*n+15)/(1210928732*n^4+8646713647*n^3+22366953268*n^2+24525178373*n+9375085020)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 1, 3, 10, 30, 90, 273, 840, 2628, 8325, 26620, 85668, 276926, 898079, 2919840, 9513000, 31051112, 101521467, 332425311, 1090002450, 3578531460, 11761905771, 38699236224, 127450069464, 420103593100, 1385865433875, 4575170995551, 15114420231498, 49963452087220] Doing the same for vertices with 5 children, we obtain the recurrence b[n+5] = 1/8*(1244834736496*n^8+24765155229984*n^7+207413359877056*n^6+944788305027984*n^5+2516456700356704*n^4+3894433596605616*n^3+3212225074504944*n^2+1073352912611616*n)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n]+1/8*(400817410292*n^8+8174403239314*n^7+71257407389996*n^6+344359288233124*n^5+1000462686324062*n^4+1770431550273736*n^3+1838879323177254*n^2+1008695371051026*n+220660378495596)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+1]+1/8*(-686293858861*n^8-14339651569555*n^7-127548986663355*n^6-627006198262910*n^5-1845012386188568*n^4-3271346341617045*n^3-3297833186537970*n^2-1582899040284840*n-185885259330096)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+2]+1/8*(-290622895680*n^8-6217682646240*n^7-56224087336140*n^6-276429741085590*n^5-785699114152410*n^4-1239267451612170*n^3-845915724725850*n^2+152651419215480*n+376240288473000)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+3]+1/8*(387497194240*n^8+8483992125440*n^7+77876055285040*n^6+384789701860400*n^5+1081634436476560*n^4+1623077132697020*n^3+857502464402100*n^2-672067151983440*n-778708839933360)/(n-1)/(2*n+9)/(4*n+15)/(4*n+21)/(302732183*n^4+2995332652*n^3+11070978817*n^2+18099320987*n+11036536164)*b[n+4] and the sequence [0, 0, 0, 0, 0, 1, 6, 28, 112, 414, 1462, 5027, 17028, 57200, 191282, 638106, 2125776, 7075944, 23540580, 78286802, 260284876, 865243071, 2876003438, 9559397287, 31774891368, 105625297520, 351152530820, 1167556333320, 3882589502820, 12913142204865] Armed with this information, we can compute both the expectation E[X_2] = 6.053041 and E[X_5] = 1.564419. In order to calculate the variance of X_2 and X_5, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[2]*d/dy[2] twice rather than once. Doing so gives Var(X_2)= 4.025753, and making a similar calculation for 5 gives Var(X_5)= 0.609571. With these calculations finished, we begin demonstrating the fact that X_2 and X_5 are jointly normally distributed. In order to do so, we calculate E((X[2]-6.053040808)^p1*(X[5]-1.564419059)^p2/(4.025752977^(1/2*p1))/(.6095706484^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[2]*d/dy[2] p1 times and y[5]*d/dy[5] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_2 and X_5 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_2 and X_5 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_2 and X_5. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_2 and X_5 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_2 and X_5 is -0.796845, while the moment would be -0.796845 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_2 and X_5 is -0.009677, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_2 and X_5 is -2.208909, while the moment would be -2.390535 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_2 and X_5 is -0.402533, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_2 and X_5 is -0.068876, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_2 and X_5 is 2.098087, while the moment would be 2.269924 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_2 and X_5 is 0.035551, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_2 and X_5 is 8.256186, while the moment would be 10.619544 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_2 and X_5 is -2.222790, while the moment would be -2.390535 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_2 and X_5 is 0.286742, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_2 and X_5 is -7.946410, while the moment would be -10.207404 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_2 and X_5 is -0.949609, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_2 and X_5 is -0.650275, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_2 and X_5 is 8.394285, while the moment would be 10.619544 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_2 and X_5 is -0.942255, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_2 and X_5 is 37.269226, while the moment would be 64.393509 if the random variables were normally distributed. On Statistics of Ordered Trees Where the Number of Children a Node Can Have is Restricted to {0, 1, 3, 4}. By Lumpy (AKA Yonah's Computer) This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 2/3*(19278*n^6+168399*n^5+547074*n^4+810891*n^3+537354*n^2+124416*n)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n]+2/3*(-10710*n^6-109620*n^5-444357*n^4-909144*n^3-986895*n^2-536598*n-113400)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+1]+2/3*(6664*n^6+78204*n^5+370615*n^4+903426*n^3+1186885*n^2+789174*n+204120)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_1, the numbers of vertices with 0 and 1 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[4])+y[3]*f(x,y[0],y[1],y[3],y[4])^3+y[4]*f(x,y[0],y[1],y[3],y[4])^4). Our next step is to compute E[X_0] and E[X_1]. We do this by finding the total number of vertices with 0 and 1 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 1. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+3] = 2/3*(38556*n^6+155358*n^5+194400*n^4+48519*n^3-50949*n^2-21870*n)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n]+2/3*(-21420*n^6-97020*n^5-153990*n^4-93348*n^3-4815*n^2+13221*n+3078)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n+1]+2/3*(13328*n^6+67032*n^5+120092*n^4+82725*n^3+2390*n^2-18429*n-4698)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 1, 4, 17, 51, 136, 393, 1233, 3865, 11851, 36301, 112520, 351352, 1098189, 3433704, 10758609, 33794505, 106344793, 335061790, 1056924667, 3338026857, 10554163533, 33402840615, 105809430024, 335444908176, 1064268538776, 3379009937161, 10735253448349, 34127137228747] Doing the same for vertices with 1 children, we obtain the recurrence b[n+3] = 2/3*(19278*n^6+91287*n^5+138267*n^4+53541*n^3-33129*n^2-20412*n)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n]+2/3*(-10710*n^6-56070*n^5-102267*n^4-70623*n^3-3771*n^2+13293*n+3348)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n+1]+2/3*(6664*n^6+38220*n^5+79555*n^4+69726*n^3+18217*n^2-5742*n-2520)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n+2] and the sequence [0, 1, 2, 3, 8, 30, 96, 273, 792, 2439, 7630, 23606, 72744, 225667, 704718, 2205855, 6909984, 21679641, 68159682, 214662589, 676924040, 2137021677, 6753952062, 21367885914, 67666240512, 214460832600, 680239606176, 2159195043357, 6858287550688, 21797761822658] Armed with this information, we can compute both the expectation E[X_0] = 14.769283 and E[X_1] = 9.433470. In order to calculate the variance of X_0 and X_1, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.354454, and making a similar calculation for 1 gives Var(X_1)= 6.517368. With these calculations finished, we begin demonstrating the fact that X_0 and X_1 are jointly normally distributed. In order to do so, we calculate E((X[0]-14.76928310)^p1*(X[1]-9.433469708)^p2/(3.354453663^(1/2*p1))/(6.517368390^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[1]*d/dy[1] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_1 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_1 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_1. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_1 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_1 is -0.982497, while the moment would be -0.982497 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_1 is -0.147006, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_1 is -2.905320, while the moment would be -2.947490 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_1 is -1.432870, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_1 is 0.139913, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_1 is 2.888435, while the moment would be 2.930600 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_1 is 1.399750, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_1 is 14.171253, while the moment would be 14.583597 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_1 is -2.904697, while the moment would be -2.947490 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_1 is -1.359149, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_1 is -14.112049, while the moment would be -14.532893 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_1 is -13.949963, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_1 is 1.310290, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_1 is 14.146207, while the moment would be 14.583597 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_1 is 13.653089, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_1 is 97.588011, while the moment would be 100.864869 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 2/3*(19278*n^6+168399*n^5+547074*n^4+810891*n^3+537354*n^2+124416*n)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n]+2/3*(-10710*n^6-109620*n^5-444357*n^4-909144*n^3-986895*n^2-536598*n-113400)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+1]+2/3*(6664*n^6+78204*n^5+370615*n^4+903426*n^3+1186885*n^2+789174*n+204120)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_3, the numbers of vertices with 0 and 3 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[4])+y[3]*f(x,y[0],y[1],y[3],y[4])^3+y[4]*f(x,y[0],y[1],y[3],y[4])^4). Our next step is to compute E[X_0] and E[X_3]. We do this by finding the total number of vertices with 0 and 3 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 3. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+3] = 2/3*(38556*n^6+155358*n^5+194400*n^4+48519*n^3-50949*n^2-21870*n)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n]+2/3*(-21420*n^6-97020*n^5-153990*n^4-93348*n^3-4815*n^2+13221*n+3078)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n+1]+2/3*(13328*n^6+67032*n^5+120092*n^4+82725*n^3+2390*n^2-18429*n-4698)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 1, 4, 17, 51, 136, 393, 1233, 3865, 11851, 36301, 112520, 351352, 1098189, 3433704, 10758609, 33794505, 106344793, 335061790, 1056924667, 3338026857, 10554163533, 33402840615, 105809430024, 335444908176, 1064268538776, 3379009937161, 10735253448349, 34127137228747] Doing the same for vertices with 3 children, we obtain the recurrence b[n+3] = 2/3*(n+2)*(19278*n^6+118503*n^5+239355*n^4+147825*n^3-44793*n^2-52488*n)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n]+2/3*(n+2)*(-10710*n^6-71190*n^5-153702*n^4-80370*n^3+119700*n^2+151560*n+44712)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n+1]+2/3*(n+2)*(6664*n^6+47628*n^5+111202*n^4+62868*n^3-95942*n^2-112980*n-19440)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n+2] and the sequence [0, 0, 0, 1, 4, 10, 26, 84, 280, 876, 2670, 8272, 26048, 82082, 257712, 809445, 2549808, 8048752, 25427580, 80383395, 254357180, 805681548, 2554185942, 8103128022, 25724489472, 81719749800, 259762343230, 826176614445, 2629055981916, 8370354658134] Armed with this information, we can compute both the expectation E[X_0] = 14.769283 and E[X_3] = 3.622458. In order to calculate the variance of X_0 and X_3, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.354454, and making a similar calculation for 3 gives Var(X_3)= 2.074838. With these calculations finished, we begin demonstrating the fact that X_0 and X_3 are jointly normally distributed. In order to do so, we calculate E((X[0]-14.76928310)^p1*(X[3]-3.622458478)^p2/(3.354453663^(1/2*p1))/(2.074837962^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[3]*d/dy[3] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_3 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_3 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_3. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_3 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_3 is 0.137890, while the moment would be 0.137890 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_3 is 0.090524, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_3 is 0.444566, while the moment would be 0.413670 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_3 is 0.555954, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_3 is -0.091011, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_3 is 1.012303, while the moment would be 1.038027 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_3 is -0.107253, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_3 is 2.881670, while the moment would be 3.228164 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_3 is 0.411903, while the moment would be 0.413670 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_3 is 0.064411, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_3 is 1.220029, while the moment would be 1.256740 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_3 is 0.850593, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_3 is -0.600220, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_3 is 3.028977, while the moment would be 3.228164 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_3 is -1.111020, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_3 is 8.818235, while the moment would be 10.377659 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 2/3*(19278*n^6+168399*n^5+547074*n^4+810891*n^3+537354*n^2+124416*n)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n]+2/3*(-10710*n^6-109620*n^5-444357*n^4-909144*n^3-986895*n^2-536598*n-113400)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+1]+2/3*(6664*n^6+78204*n^5+370615*n^4+903426*n^3+1186885*n^2+789174*n+204120)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_4, the numbers of vertices with 0 and 4 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[4])+y[3]*f(x,y[0],y[1],y[3],y[4])^3+y[4]*f(x,y[0],y[1],y[3],y[4])^4). Our next step is to compute E[X_0] and E[X_4]. We do this by finding the total number of vertices with 0 and 4 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 4. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+3] = 2/3*(38556*n^6+155358*n^5+194400*n^4+48519*n^3-50949*n^2-21870*n)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n]+2/3*(-21420*n^6-97020*n^5-153990*n^4-93348*n^3-4815*n^2+13221*n+3078)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n+1]+2/3*(13328*n^6+67032*n^5+120092*n^4+82725*n^3+2390*n^2-18429*n-4698)/(3*n+5)/(3*n+4)/(n+2)/(238*n^3+126*n^2-67*n-27)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 1, 4, 17, 51, 136, 393, 1233, 3865, 11851, 36301, 112520, 351352, 1098189, 3433704, 10758609, 33794505, 106344793, 335061790, 1056924667, 3338026857, 10554163533, 33402840615, 105809430024, 335444908176, 1064268538776, 3379009937161, 10735253448349, 34127137228747] Doing the same for vertices with 4 children, we obtain the recurrence b[n+3] = 2/3*(n+2)*(5508*n^6+37746*n^5+85536*n^4+61641*n^3-13527*n^2-21870*n)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n]+2/3*(n+2)*(-3060*n^6-22500*n^5-51390*n^4-15480*n^3+87525*n^2+106425*n+35370)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n+1]+2/3*(n+2)*(1904*n^6+14952*n^5+35240*n^4+735*n^3-99784*n^2-113517*n-29070)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n+2] and the sequence [0, 0, 0, 0, 1, 5, 15, 42, 134, 450, 1455, 4565, 14355, 45617, 145236, 460980, 1461240, 4636784, 14730534, 46816266, 148815790, 473221980, 1505584377, 4792385961, 15260672592, 48614115600, 154924555175, 493905955725, 1575164841705, 5025248172141] Armed with this information, we can compute both the expectation E[X_0] = 14.769283 and E[X_4] = 2.174789. In order to calculate the variance of X_0 and X_4, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.354454, and making a similar calculation for 4 gives Var(X_4)= 1.133188. With these calculations finished, we begin demonstrating the fact that X_0 and X_4 are jointly normally distributed. In order to do so, we calculate E((X[0]-14.76928310)^p1*(X[4]-2.174788714)^p2/(3.354453663^(1/2*p1))/(1.133188449^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[4]*d/dy[4] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_4 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_4 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_4. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_4 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_4 is 0.449118, while the moment would be 0.449118 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_4 is 0.125612, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_4 is 1.281853, while the moment would be 1.347353 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_4 is 0.925348, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_4 is 0.008478, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_4 is 1.369706, while the moment would be 1.403413 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_4 is 0.467068, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_4 is 4.835978, while the moment would be 5.420480 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_4 is 1.323485, while the moment would be 1.347353 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_4 is 0.261756, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_4 is 4.183445, while the moment would be 4.585599 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_4 is 3.212830, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_4 is -0.176453, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_4 is 5.029494, while the moment would be 5.420480 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_4 is 1.740850, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_4 is 20.355815, while the moment would be 24.499335 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 2/3*(19278*n^6+168399*n^5+547074*n^4+810891*n^3+537354*n^2+124416*n)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n]+2/3*(-10710*n^6-109620*n^5-444357*n^4-909144*n^3-986895*n^2-536598*n-113400)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+1]+2/3*(6664*n^6+78204*n^5+370615*n^4+903426*n^3+1186885*n^2+789174*n+204120)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_3, the numbers of vertices with 1 and 3 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[4])+y[3]*f(x,y[0],y[1],y[3],y[4])^3+y[4]*f(x,y[0],y[1],y[3],y[4])^4). Our next step is to compute E[X_1] and E[X_3]. We do this by finding the total number of vertices with 1 and 3 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 3. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+3] = 2/3*(19278*n^6+91287*n^5+138267*n^4+53541*n^3-33129*n^2-20412*n)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n]+2/3*(-10710*n^6-56070*n^5-102267*n^4-70623*n^3-3771*n^2+13293*n+3348)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n+1]+2/3*(6664*n^6+38220*n^5+79555*n^4+69726*n^3+18217*n^2-5742*n-2520)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 3, 8, 30, 96, 273, 792, 2439, 7630, 23606, 72744, 225667, 704718, 2205855, 6909984, 21679641, 68159682, 214662589, 676924040, 2137021677, 6753952062, 21367885914, 67666240512, 214460832600, 680239606176, 2159195043357, 6858287550688, 21797761822658] Doing the same for vertices with 3 children, we obtain the recurrence b[n+3] = 2/3*(n+2)*(19278*n^6+118503*n^5+239355*n^4+147825*n^3-44793*n^2-52488*n)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n]+2/3*(n+2)*(-10710*n^6-71190*n^5-153702*n^4-80370*n^3+119700*n^2+151560*n+44712)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n+1]+2/3*(n+2)*(6664*n^6+47628*n^5+111202*n^4+62868*n^3-95942*n^2-112980*n-19440)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n+2] and the sequence [0, 0, 0, 1, 4, 10, 26, 84, 280, 876, 2670, 8272, 26048, 82082, 257712, 809445, 2549808, 8048752, 25427580, 80383395, 254357180, 805681548, 2554185942, 8103128022, 25724489472, 81719749800, 259762343230, 826176614445, 2629055981916, 8370354658134] Armed with this information, we can compute both the expectation E[X_1] = 9.433470 and E[X_3] = 3.622458. In order to calculate the variance of X_1 and X_3, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.517368, and making a similar calculation for 3 gives Var(X_3)= 2.074838. With these calculations finished, we begin demonstrating the fact that X_1 and X_3 are jointly normally distributed. In order to do so, we calculate E((X[1]-9.433469708)^p1*(X[3]-3.622458478)^p2/(6.517368390^(1/2*p1))/(2.074837962^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[3]*d/dy[3] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_3 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_3 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_3. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_3 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_3 is -0.319977, while the moment would be -0.319977 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_3 is -0.101563, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_3 is -0.940900, while the moment would be -0.959931 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_3 is -0.665177, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_3 is -0.047889, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_3 is 1.183211, while the moment would be 1.204771 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_3 is 0.126979, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_3 is 3.839215, while the moment would be 4.228623 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_3 is -0.946522, while the moment would be -0.959931 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_3 is -0.062143, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_3 is -2.856680, while the moment would be -3.076359 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_3 is -1.359792, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_3 is -0.465949, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_3 is 3.969836, while the moment would be 4.228623 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_3 is -0.196981, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_3 is 14.181186, while the moment would be 16.623326 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 2/3*(19278*n^6+168399*n^5+547074*n^4+810891*n^3+537354*n^2+124416*n)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n]+2/3*(-10710*n^6-109620*n^5-444357*n^4-909144*n^3-986895*n^2-536598*n-113400)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+1]+2/3*(6664*n^6+78204*n^5+370615*n^4+903426*n^3+1186885*n^2+789174*n+204120)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_4, the numbers of vertices with 1 and 4 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[4])+y[3]*f(x,y[0],y[1],y[3],y[4])^3+y[4]*f(x,y[0],y[1],y[3],y[4])^4). Our next step is to compute E[X_1] and E[X_4]. We do this by finding the total number of vertices with 1 and 4 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 4. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+3] = 2/3*(19278*n^6+91287*n^5+138267*n^4+53541*n^3-33129*n^2-20412*n)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n]+2/3*(-10710*n^6-56070*n^5-102267*n^4-70623*n^3-3771*n^2+13293*n+3348)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n+1]+2/3*(6664*n^6+38220*n^5+79555*n^4+69726*n^3+18217*n^2-5742*n-2520)/(n+1)/(3*n+5)/(119*n^3+147*n^2+10*n-24)/(3*n+7)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 3, 8, 30, 96, 273, 792, 2439, 7630, 23606, 72744, 225667, 704718, 2205855, 6909984, 21679641, 68159682, 214662589, 676924040, 2137021677, 6753952062, 21367885914, 67666240512, 214460832600, 680239606176, 2159195043357, 6858287550688, 21797761822658] Doing the same for vertices with 4 children, we obtain the recurrence b[n+3] = 2/3*(n+2)*(5508*n^6+37746*n^5+85536*n^4+61641*n^3-13527*n^2-21870*n)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n]+2/3*(n+2)*(-3060*n^6-22500*n^5-51390*n^4-15480*n^3+87525*n^2+106425*n+35370)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n+1]+2/3*(n+2)*(1904*n^6+14952*n^5+35240*n^4+735*n^3-99784*n^2-113517*n-29070)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n+2] and the sequence [0, 0, 0, 0, 1, 5, 15, 42, 134, 450, 1455, 4565, 14355, 45617, 145236, 460980, 1461240, 4636784, 14730534, 46816266, 148815790, 473221980, 1505584377, 4792385961, 15260672592, 48614115600, 154924555175, 493905955725, 1575164841705, 5025248172141] Armed with this information, we can compute both the expectation E[X_1] = 9.433470 and E[X_4] = 2.174789. In order to calculate the variance of X_1 and X_4, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.517368, and making a similar calculation for 4 gives Var(X_4)= 1.133188. With these calculations finished, we begin demonstrating the fact that X_1 and X_4 are jointly normally distributed. In order to do so, we calculate E((X[1]-9.433469708)^p1*(X[4]-2.174788714)^p2/(6.517368390^(1/2*p1))/(1.133188449^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[4]*d/dy[4] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_4 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_4 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_4. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_4 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_4 is -0.274821, while the moment would be -0.274821 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_4 is -0.109093, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_4 is -0.809853, while the moment would be -0.824462 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_4 is -0.744539, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_4 is -0.041109, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_4 is 1.129758, while the moment would be 1.151053 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_4 is 0.178049, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_4 is 3.535355, while the moment would be 3.906316 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_4 is -0.812372, while the moment would be -0.824462 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_4 is -0.106832, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_4 is -2.408446, while the moment would be -2.597922 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_4 is -1.644710, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_4 is -0.401433, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_4 is 3.656536, while the moment would be 3.906316 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_4 is 0.104916, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_4 is 12.223660, while the moment would be 14.574798 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^4; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^4+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+3] = 2/3*(19278*n^6+168399*n^5+547074*n^4+810891*n^3+537354*n^2+124416*n)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n]+2/3*(-10710*n^6-109620*n^5-444357*n^4-909144*n^3-986895*n^2-536598*n-113400)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+1]+2/3*(6664*n^6+78204*n^5+370615*n^4+903426*n^3+1186885*n^2+789174*n+204120)/(3*n+10)/(n+3)/(3*n+8)/(119*n^3+504*n^2+661*n+252)*b[n+2]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 6, 16, 39, 99, 271, 763, 2146, 6062, 17359, 50337, 147057, 431874, 1275273, 3786649, 11298031, 33846202, 101762937, 306997821, 929038518, 2819426688, 8578433304, 26163061776, 79970186791, 244938841096, 751646959402, 2310683396056] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_3 and X_4, the numbers of vertices with 3 and 4 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[4]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[4])+y[3]*f(x,y[0],y[1],y[3],y[4])^3+y[4]*f(x,y[0],y[1],y[3],y[4])^4). Our next step is to compute E[X_3] and E[X_4]. We do this by finding the total number of vertices with 3 and 4 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[3]*d/dy[3] to multiply every monomial by the number of nodes with 3 children in the corresponding tree, and similarly for 4. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 3 children over all trees with n vertices: b[n+3] = 2/3*(n+2)*(19278*n^6+118503*n^5+239355*n^4+147825*n^3-44793*n^2-52488*n)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n]+2/3*(n+2)*(-10710*n^6-71190*n^5-153702*n^4-80370*n^3+119700*n^2+151560*n+44712)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n+1]+2/3*(n+2)*(6664*n^6+47628*n^5+111202*n^4+62868*n^3-95942*n^2-112980*n-19440)/n/(n-1)/(3*n+8)/(3*n+7)/(n+3)/(119*n^2+315*n+214)*b[n+2]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 0, 1, 4, 10, 26, 84, 280, 876, 2670, 8272, 26048, 82082, 257712, 809445, 2549808, 8048752, 25427580, 80383395, 254357180, 805681548, 2554185942, 8103128022, 25724489472, 81719749800, 259762343230, 826176614445, 2629055981916, 8370354658134] Doing the same for vertices with 4 children, we obtain the recurrence b[n+3] = 2/3*(n+2)*(5508*n^6+37746*n^5+85536*n^4+61641*n^3-13527*n^2-21870*n)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n]+2/3*(n+2)*(-3060*n^6-22500*n^5-51390*n^4-15480*n^3+87525*n^2+106425*n+35370)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n+1]+2/3*(n+2)*(1904*n^6+14952*n^5+35240*n^4+735*n^3-99784*n^2-113517*n-29070)/(n-2)/(3*n+8)/(3*n+10)/(n+3)/(34*n^3+114*n^2+107*n+15)*b[n+2] and the sequence [0, 0, 0, 0, 1, 5, 15, 42, 134, 450, 1455, 4565, 14355, 45617, 145236, 460980, 1461240, 4636784, 14730534, 46816266, 148815790, 473221980, 1505584377, 4792385961, 15260672592, 48614115600, 154924555175, 493905955725, 1575164841705, 5025248172141] Armed with this information, we can compute both the expectation E[X_3] = 3.622458 and E[X_4] = 2.174789. In order to calculate the variance of X_3 and X_4, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[3]*d/dy[3] twice rather than once. Doing so gives Var(X_3)= 2.074838, and making a similar calculation for 4 gives Var(X_4)= 1.133188. With these calculations finished, we begin demonstrating the fact that X_3 and X_4 are jointly normally distributed. In order to do so, we calculate E((X[3]-3.622458478)^p1*(X[4]-2.174788714)^p2/(2.074837962^(1/2*p1))/(1.133188449^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[3]*d/dy[3] p1 times and y[4]*d/dy[4] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_3 and X_4 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_3 and X_4 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_3 and X_4. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_3 and X_4 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_3 and X_4 is -0.823009, while the moment would be -0.823009 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_3 and X_4 is -0.058820, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_3 and X_4 is -2.213567, while the moment would be -2.469027 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_3 and X_4 is -0.747640, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_3 and X_4 is -0.019895, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_3 and X_4 is 2.104038, while the moment would be 2.354687 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_3 and X_4 is 0.365423, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_3 and X_4 is 7.999135, while the moment would be 11.128125 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_3 and X_4 is -2.218272, while the moment would be -2.469027 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_3 and X_4 is -0.033437, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_3 and X_4 is -7.662435, while the moment would be -10.751840 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_3 and X_4 is -2.706635, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_3 and X_4 is -0.320883, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_3 and X_4 is 7.998671, while the moment would be 11.128125 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_3 and X_4 is 0.866408, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_3 and X_4 is 34.923579, while the moment would be 68.779817 if the random variables were normally distributed. On Statistics of Ordered Trees Where the Number of Children a Node Can Have is Restricted to {0, 1, 3, 5}. By Lumpy (AKA Yonah's Computer) This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(86367275456*n^8+1771089488168*n^7+15173633264284*n^6+70215512147462*n^5+188830918880329*n^4+293557555532912*n^3+242316279227231*n^2+80863041478758*n)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n]+1/8*(-75544840448*n^8-1662476616416*n^7-15597026735488*n^6-81348027774904*n^5-257463090096696*n^4-505107996702144*n^3-598117449257592*n^2-389495949651336*n-106361339814576)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+1]+1/8*(74078699520*n^8+1741330053120*n^7+17631543383424*n^6+100424836970608*n^5+351838674172528*n^4+776146480665380*n^3+1052288869518716*n^2+801178406443872*n+262056499618032)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+2]+1/8*(-39817300992*n^8-995690855040*n^7-10771193872128*n^6-65816150217856*n^5-248327857631632*n^4-592000367818040*n^3-869916296907092*n^2-719387837821524*n-255829062547296)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+3]+1/8*(22223609856*n^8+589069845504*n^7+6751640169216*n^6+43665729800224*n^5+174100330761904*n^4+437602162116296*n^3+675924232731524*n^2+585195750853356*n+216713078305920)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 5, 12, 30, 78, 205, 547, 1488, 4104, 11430, 32111, 90910, 259087, 742701, 2140165, 6195787, 18011450, 52556756, 153881305, 451948784, 1331141032, 3930882046, 11635856286, 34519968211, 102620956036, 305655819202, 912019030930] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_1, the numbers of vertices with 0 and 1 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[5])+y[3]*f(x,y[0],y[1],y[3],y[5])^3+y[5]*f(x,y[0],y[1],y[3],y[5])^5). Our next step is to compute E[X_0] and E[X_1]. We do this by finding the total number of vertices with 0 and 1 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 1. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+5] = 1/8*(86367275456*n^8+1335891062968*n^7+8640053473084*n^6+30230691367162*n^5+61655302669829*n^4+73096922554612*n^3+46514854800931*n^2+12233073234558*n)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n]+1/8*(-75544840448*n^8-1206266814368*n^7-8136546394944*n^6-30130796131832*n^5-66533482395600*n^4-88789191303912*n^3-68799875244496*n^2-27493314224288*n-4074119598912)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+1]+1/8*(74078699520*n^8+1219895470080*n^7+8513464495744*n^6+32730948740528*n^5+75295863561328*n^4+104980768664660*n^3+85014310541956*n^2+35193212564792*n+5152620360192)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+2]+1/8*(-39817300992*n^8-675602465664*n^7-4868477921024*n^6-19360640975360*n^5-46117971998768*n^4-66573043349656*n^3-55685291090516*n^2-23629947789780*n-3455295279840)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+3]+1/8*(22223609856*n^8+388192250880*n^7+2870642576128*n^6+11664830631136*n^5+28229035586544*n^4+41062745586040*n^3+34167112335332*n^2+14056099566844*n+1830901897440)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 1, 4, 13, 36, 106, 316, 925, 2731, 8176, 24586, 74086, 223939, 678770, 2061879, 6275949, 19137989, 58453243, 178787284, 547549468, 1678872073, 5153155228, 15832542408, 48687259534, 149843012686, 461516192411, 1422468062536, 4387141573114, 13538948057158] Doing the same for vertices with 1 children, we obtain the recurrence b[n+5] = 1/8*(n+4)*(86367275456*n^8+1425620386344*n^7+9860364799780*n^6+36973260867158*n^5+80978768333727*n^4+103292972104564*n^3+70800582548553*n^2+20034229599450*n)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n]+1/8*(n+4)*(-75544840448*n^8-1284752414176*n^7-9249299631616*n^6-36672203100984*n^5-87121905236168*n^4-126090978398000*n^3-107505019194760*n^2-48674745715992*n-8770910726160)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+1]+1/8*(n+4)*(74078699520*n^8+1296857856000*n^7+9665680113024*n^6+39938730293872*n^5+99653168721568*n^4+152834053511524*n^3+139400764659384*n^2+68053831603452*n+13195833777720)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+2]+1/8*(n+4)*(-39817300992*n^8-716969748096*n^7-5513760159744*n^6-23589932709760*n^5-61179358603920*n^4-97946393192456*n^3-93739254528228*n^2-48353658166512*n-10026272320020)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+3]+1/8*(n+4)*(22223609856*n^8+411280966656*n^7+3250412326656*n^6+14281833388576*n^5+37984492396304*n^4+62198256127304*n^3+60588829810444*n^2+31516956717084*n+6458792963040)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+4] and the sequence [0, 1, 2, 3, 8, 25, 72, 210, 624, 1845, 5470, 16368, 49248, 148590, 449554, 1363650, 4145392, 12625917, 38522970, 117719953, 360229000, 1103691876, 3385388710, 10394822032, 31947384768, 98272051150, 302532263436, 932039141697, 2873386769008, 8864018756858] Armed with this information, we can compute both the expectation E[X_0] = 14.845028 and E[X_1] = 9.719116. In order to calculate the variance of X_0 and X_1, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.769367, and making a similar calculation for 1 gives Var(X_1)= 6.624652. With these calculations finished, we begin demonstrating the fact that X_0 and X_1 are jointly normally distributed. In order to do so, we calculate E((X[0]-14.84502801)^p1*(X[1]-9.719116002)^p2/(3.769367454^(1/2*p1))/(6.624651810^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[1]*d/dy[1] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_1 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_1 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_1. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_1 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_1 is -0.956571, while the moment would be -0.956571 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_1 is -0.135050, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_1 is -2.826176, while the moment would be -2.869712 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_1 is -1.317877, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_1 is 0.117899, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_1 is 2.786816, while the moment would be 2.830055 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_1 is 1.239317, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_1 is 13.521806, while the moment would be 13.980332 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_1 is -2.825437, while the moment would be -2.869712 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_1 is -1.142362, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_1 is -13.385199, while the moment would be -13.860869 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_1 is -12.022962, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_1 is 1.022403, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_1 is 13.469319, while the moment would be 13.980332 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_1 is 11.341003, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_1 is 90.482674, while the moment would be 94.976605 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(86367275456*n^8+1771089488168*n^7+15173633264284*n^6+70215512147462*n^5+188830918880329*n^4+293557555532912*n^3+242316279227231*n^2+80863041478758*n)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n]+1/8*(-75544840448*n^8-1662476616416*n^7-15597026735488*n^6-81348027774904*n^5-257463090096696*n^4-505107996702144*n^3-598117449257592*n^2-389495949651336*n-106361339814576)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+1]+1/8*(74078699520*n^8+1741330053120*n^7+17631543383424*n^6+100424836970608*n^5+351838674172528*n^4+776146480665380*n^3+1052288869518716*n^2+801178406443872*n+262056499618032)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+2]+1/8*(-39817300992*n^8-995690855040*n^7-10771193872128*n^6-65816150217856*n^5-248327857631632*n^4-592000367818040*n^3-869916296907092*n^2-719387837821524*n-255829062547296)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+3]+1/8*(22223609856*n^8+589069845504*n^7+6751640169216*n^6+43665729800224*n^5+174100330761904*n^4+437602162116296*n^3+675924232731524*n^2+585195750853356*n+216713078305920)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 5, 12, 30, 78, 205, 547, 1488, 4104, 11430, 32111, 90910, 259087, 742701, 2140165, 6195787, 18011450, 52556756, 153881305, 451948784, 1331141032, 3930882046, 11635856286, 34519968211, 102620956036, 305655819202, 912019030930] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_3, the numbers of vertices with 0 and 3 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[5])+y[3]*f(x,y[0],y[1],y[3],y[5])^3+y[5]*f(x,y[0],y[1],y[3],y[5])^5). Our next step is to compute E[X_0] and E[X_3]. We do this by finding the total number of vertices with 0 and 3 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 3. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+5] = 1/8*(86367275456*n^8+1335891062968*n^7+8640053473084*n^6+30230691367162*n^5+61655302669829*n^4+73096922554612*n^3+46514854800931*n^2+12233073234558*n)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n]+1/8*(-75544840448*n^8-1206266814368*n^7-8136546394944*n^6-30130796131832*n^5-66533482395600*n^4-88789191303912*n^3-68799875244496*n^2-27493314224288*n-4074119598912)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+1]+1/8*(74078699520*n^8+1219895470080*n^7+8513464495744*n^6+32730948740528*n^5+75295863561328*n^4+104980768664660*n^3+85014310541956*n^2+35193212564792*n+5152620360192)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+2]+1/8*(-39817300992*n^8-675602465664*n^7-4868477921024*n^6-19360640975360*n^5-46117971998768*n^4-66573043349656*n^3-55685291090516*n^2-23629947789780*n-3455295279840)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+3]+1/8*(22223609856*n^8+388192250880*n^7+2870642576128*n^6+11664830631136*n^5+28229035586544*n^4+41062745586040*n^3+34167112335332*n^2+14056099566844*n+1830901897440)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 1, 4, 13, 36, 106, 316, 925, 2731, 8176, 24586, 74086, 223939, 678770, 2061879, 6275949, 19137989, 58453243, 178787284, 547549468, 1678872073, 5153155228, 15832542408, 48687259534, 149843012686, 461516192411, 1422468062536, 4387141573114, 13538948057158] Doing the same for vertices with 3 children, we obtain the recurrence b[n+5] = 1/8*(n+4)*(43183637728*n^8+802539516548*n^7+6250574393294*n^6+26383956008773*n^5+64984279519115*n^4+93009471856125*n^3+71205660425119*n^2+22287730593810*n)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n]+1/8*(n+4)*(-37772420224*n^8-720861806896*n^7-5824942898544*n^6-25937291880060*n^5-69285015591572*n^4-113009667340680*n^3-109072005282372*n^2-56421450343676*n-11869535178600)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+1]+1/8*(n+4)*(37039349760*n^8+725391313920*n^7+6058800922496*n^6+28119836193280*n^5+79047409894388*n^4+137152613281148*n^3+142387742590212*n^2+79890265297748*n+18131421962040)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+2]+1/8*(n+4)*(-19908650496*n^8-399852156480*n^7-3430013629504*n^6-16371160485472*n^5-47381234146192*n^4-84727447649764*n^3-90781694559188*n^2-52738164303084*n-12508434957420)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+3]+1/8*(n+4)*(11111804928*n^8+228729199104*n^7+2008041585536*n^6+9790512953360*n^5+28885142347272*n^4+52544792647716*n^3+57175977965824*n^2+33718519987780*n+8142942662400)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+4] and the sequence [0, 0, 0, 1, 4, 10, 26, 77, 232, 696, 2100, 6347, 19184, 58136, 176722, 538330, 1642528, 5019216, 15358752, 47055609, 144330460, 443154138, 1361952966, 4189339968, 12896624952, 39730821000, 122483259600, 377835576300, 1166237952516, 3601743714654] Armed with this information, we can compute both the expectation E[X_0] = 14.845028 and E[X_3] = 3.949198. In order to calculate the variance of X_0 and X_3, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.769367, and making a similar calculation for 3 gives Var(X_3)= 2.256604. With these calculations finished, we begin demonstrating the fact that X_0 and X_3 are jointly normally distributed. In order to do so, we calculate E((X[0]-14.84502801)^p1*(X[3]-3.949197980)^p2/(3.769367454^(1/2*p1))/(2.256604355^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[3]*d/dy[3] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_3 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_3 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_3. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_3 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_3 is 0.046868, while the moment would be 0.046868 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_3 is 0.096774, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_3 is 0.210783, while the moment would be 0.140605 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_3 is 0.562658, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_3 is -0.116752, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_3 is 0.965726, while the moment would be 1.004393 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_3 is -0.266234, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_3 is 2.599220, while the moment would be 3.026360 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_3 is 0.147152, while the moment would be 0.140605 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_3 is 0.158024, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_3 is 0.517776, while the moment would be 0.422432 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_3 is 1.045487, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_3 is -0.697380, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_3 is 2.802316, while the moment would be 3.026360 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_3 is -1.656392, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_3 is 7.577448, while the moment would be 9.158273 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(86367275456*n^8+1771089488168*n^7+15173633264284*n^6+70215512147462*n^5+188830918880329*n^4+293557555532912*n^3+242316279227231*n^2+80863041478758*n)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n]+1/8*(-75544840448*n^8-1662476616416*n^7-15597026735488*n^6-81348027774904*n^5-257463090096696*n^4-505107996702144*n^3-598117449257592*n^2-389495949651336*n-106361339814576)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+1]+1/8*(74078699520*n^8+1741330053120*n^7+17631543383424*n^6+100424836970608*n^5+351838674172528*n^4+776146480665380*n^3+1052288869518716*n^2+801178406443872*n+262056499618032)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+2]+1/8*(-39817300992*n^8-995690855040*n^7-10771193872128*n^6-65816150217856*n^5-248327857631632*n^4-592000367818040*n^3-869916296907092*n^2-719387837821524*n-255829062547296)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+3]+1/8*(22223609856*n^8+589069845504*n^7+6751640169216*n^6+43665729800224*n^5+174100330761904*n^4+437602162116296*n^3+675924232731524*n^2+585195750853356*n+216713078305920)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 5, 12, 30, 78, 205, 547, 1488, 4104, 11430, 32111, 90910, 259087, 742701, 2140165, 6195787, 18011450, 52556756, 153881305, 451948784, 1331141032, 3930882046, 11635856286, 34519968211, 102620956036, 305655819202, 912019030930] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_0 and X_5, the numbers of vertices with 0 and 5 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[5])+y[3]*f(x,y[0],y[1],y[3],y[5])^3+y[5]*f(x,y[0],y[1],y[3],y[5])^5). Our next step is to compute E[X_0] and E[X_5]. We do this by finding the total number of vertices with 0 and 5 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[0]*d/dy[0] to multiply every monomial by the number of nodes with 0 children in the corresponding tree, and similarly for 5. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 0 children over all trees with n vertices: b[n+5] = 1/8*(86367275456*n^8+1335891062968*n^7+8640053473084*n^6+30230691367162*n^5+61655302669829*n^4+73096922554612*n^3+46514854800931*n^2+12233073234558*n)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n]+1/8*(-75544840448*n^8-1206266814368*n^7-8136546394944*n^6-30130796131832*n^5-66533482395600*n^4-88789191303912*n^3-68799875244496*n^2-27493314224288*n-4074119598912)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+1]+1/8*(74078699520*n^8+1219895470080*n^7+8513464495744*n^6+32730948740528*n^5+75295863561328*n^4+104980768664660*n^3+85014310541956*n^2+35193212564792*n+5152620360192)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+2]+1/8*(-39817300992*n^8-675602465664*n^7-4868477921024*n^6-19360640975360*n^5-46117971998768*n^4-66573043349656*n^3-55685291090516*n^2-23629947789780*n-3455295279840)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+3]+1/8*(22223609856*n^8+388192250880*n^7+2870642576128*n^6+11664830631136*n^5+28229035586544*n^4+41062745586040*n^3+34167112335332*n^2+14056099566844*n+1830901897440)/(4*n+11)/(2*n+7)/(4*n+9)/(n+4)/(19291328*n^4+105476504*n^3+189642708*n^2+123834594*n+17159275)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [1, 1, 1, 4, 13, 36, 106, 316, 925, 2731, 8176, 24586, 74086, 223939, 678770, 2061879, 6275949, 19137989, 58453243, 178787284, 547549468, 1678872073, 5153155228, 15832542408, 48687259534, 149843012686, 461516192411, 1422468062536, 4387141573114, 13538948057158] Doing the same for vertices with 5 children, we obtain the recurrence b[n+5] = 1/8*(12338182208*n^8+254933954264*n^7+2201259221612*n^6+10263955058786*n^5+27799550332697*n^4+43495319251076*n^3+36103285414383*n^2+12102224886774*n)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n]+1/8*(-10792120064*n^8-228384973344*n^7-2040558793152*n^6-10011863043336*n^5-29343484518480*n^4-52232649467196*n^3-54641982059368*n^2-30384216311724*n-6820296304536)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+1]+1/8*(10582671360*n^8+229243914240*n^7+2109453939072*n^6+10723342056624*n^5+32726504852304*n^4+60765882731220*n^3+65797593673788*n^2+36531230915376*n+7151022214416)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+2]+1/8*(-5688185856*n^8-126062696832*n^7-1182937788672*n^6-6089182053120*n^5-18548192368944*n^4-33351086840808*n^3-32551011267108*n^2-12857613995160*n+621975064500)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+3]+1/8*(3174801408*n^8+71947975680*n^7+684275591424*n^6+3515733543648*n^5+10387329102192*n^4+16989198998520*n^3+12177653185956*n^2-1618964503188*n-5114859047640)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+4] and the sequence [0, 0, 0, 0, 0, 1, 6, 21, 64, 198, 622, 1947, 6072, 18889, 58604, 181533, 562048, 1739848, 5384988, 16666154, 51582948, 159670623, 494325128, 1530680360, 4740781896, 14686378600, 45507426250, 141043988475, 437252462220, 1355860399230] Armed with this information, we can compute both the expectation E[X_0] = 14.845028 and E[X_5] = 1.486658. In order to calculate the variance of X_0 and X_5, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[0]*d/dy[0] twice rather than once. Doing so gives Var(X_0)= 3.769367, and making a similar calculation for 5 gives Var(X_5)= 0.765564. With these calculations finished, we begin demonstrating the fact that X_0 and X_5 are jointly normally distributed. In order to do so, we calculate E((X[0]-14.84502801)^p1*(X[5]-1.486658012)^p2/(3.769367454^(1/2*p1))/(.7655637396^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[0]*d/dy[0] p1 times and y[5]*d/dy[5] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_0 and X_5 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_0 and X_5 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_0 and X_5. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_0 and X_5 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_0 and X_5 is 0.514499, while the moment would be 0.514499 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_5 is 0.155150, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_0 and X_5 is 1.438129, while the moment would be 1.543497 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_5 is 1.220792, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_0 and X_5 is 0.050906, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_0 and X_5 is 1.479611, while the moment would be 1.529419 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_5 is 0.766630, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_0 and X_5 is 5.355202, while the moment would be 6.176512 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_5 is 1.510695, while the moment would be 1.543497 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_5 is 0.513395, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_0 and X_5 is 4.881257, while the moment would be 5.447649 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_5 is 5.279646, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_0 and X_5 is 0.116928, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_0 and X_5 is 5.667114, while the moment would be 6.176512 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_5 is 3.807645, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_0 and X_5 is 24.366774, while the moment would be 29.740780 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(86367275456*n^8+1771089488168*n^7+15173633264284*n^6+70215512147462*n^5+188830918880329*n^4+293557555532912*n^3+242316279227231*n^2+80863041478758*n)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n]+1/8*(-75544840448*n^8-1662476616416*n^7-15597026735488*n^6-81348027774904*n^5-257463090096696*n^4-505107996702144*n^3-598117449257592*n^2-389495949651336*n-106361339814576)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+1]+1/8*(74078699520*n^8+1741330053120*n^7+17631543383424*n^6+100424836970608*n^5+351838674172528*n^4+776146480665380*n^3+1052288869518716*n^2+801178406443872*n+262056499618032)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+2]+1/8*(-39817300992*n^8-995690855040*n^7-10771193872128*n^6-65816150217856*n^5-248327857631632*n^4-592000367818040*n^3-869916296907092*n^2-719387837821524*n-255829062547296)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+3]+1/8*(22223609856*n^8+589069845504*n^7+6751640169216*n^6+43665729800224*n^5+174100330761904*n^4+437602162116296*n^3+675924232731524*n^2+585195750853356*n+216713078305920)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 5, 12, 30, 78, 205, 547, 1488, 4104, 11430, 32111, 90910, 259087, 742701, 2140165, 6195787, 18011450, 52556756, 153881305, 451948784, 1331141032, 3930882046, 11635856286, 34519968211, 102620956036, 305655819202, 912019030930] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_3, the numbers of vertices with 1 and 3 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[5])+y[3]*f(x,y[0],y[1],y[3],y[5])^3+y[5]*f(x,y[0],y[1],y[3],y[5])^5). Our next step is to compute E[X_1] and E[X_3]. We do this by finding the total number of vertices with 1 and 3 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 3. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+5] = 1/8*(n+4)*(86367275456*n^8+1425620386344*n^7+9860364799780*n^6+36973260867158*n^5+80978768333727*n^4+103292972104564*n^3+70800582548553*n^2+20034229599450*n)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n]+1/8*(n+4)*(-75544840448*n^8-1284752414176*n^7-9249299631616*n^6-36672203100984*n^5-87121905236168*n^4-126090978398000*n^3-107505019194760*n^2-48674745715992*n-8770910726160)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+1]+1/8*(n+4)*(74078699520*n^8+1296857856000*n^7+9665680113024*n^6+39938730293872*n^5+99653168721568*n^4+152834053511524*n^3+139400764659384*n^2+68053831603452*n+13195833777720)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+2]+1/8*(n+4)*(-39817300992*n^8-716969748096*n^7-5513760159744*n^6-23589932709760*n^5-61179358603920*n^4-97946393192456*n^3-93739254528228*n^2-48353658166512*n-10026272320020)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+3]+1/8*(n+4)*(22223609856*n^8+411280966656*n^7+3250412326656*n^6+14281833388576*n^5+37984492396304*n^4+62198256127304*n^3+60588829810444*n^2+31516956717084*n+6458792963040)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 3, 8, 25, 72, 210, 624, 1845, 5470, 16368, 49248, 148590, 449554, 1363650, 4145392, 12625917, 38522970, 117719953, 360229000, 1103691876, 3385388710, 10394822032, 31947384768, 98272051150, 302532263436, 932039141697, 2873386769008, 8864018756858] Doing the same for vertices with 3 children, we obtain the recurrence b[n+5] = 1/8*(n+4)*(43183637728*n^8+802539516548*n^7+6250574393294*n^6+26383956008773*n^5+64984279519115*n^4+93009471856125*n^3+71205660425119*n^2+22287730593810*n)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n]+1/8*(n+4)*(-37772420224*n^8-720861806896*n^7-5824942898544*n^6-25937291880060*n^5-69285015591572*n^4-113009667340680*n^3-109072005282372*n^2-56421450343676*n-11869535178600)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+1]+1/8*(n+4)*(37039349760*n^8+725391313920*n^7+6058800922496*n^6+28119836193280*n^5+79047409894388*n^4+137152613281148*n^3+142387742590212*n^2+79890265297748*n+18131421962040)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+2]+1/8*(n+4)*(-19908650496*n^8-399852156480*n^7-3430013629504*n^6-16371160485472*n^5-47381234146192*n^4-84727447649764*n^3-90781694559188*n^2-52738164303084*n-12508434957420)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+3]+1/8*(n+4)*(11111804928*n^8+228729199104*n^7+2008041585536*n^6+9790512953360*n^5+28885142347272*n^4+52544792647716*n^3+57175977965824*n^2+33718519987780*n+8142942662400)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+4] and the sequence [0, 0, 0, 1, 4, 10, 26, 77, 232, 696, 2100, 6347, 19184, 58136, 176722, 538330, 1642528, 5019216, 15358752, 47055609, 144330460, 443154138, 1361952966, 4189339968, 12896624952, 39730821000, 122483259600, 377835576300, 1166237952516, 3601743714654] Armed with this information, we can compute both the expectation E[X_1] = 9.719116 and E[X_3] = 3.949198. In order to calculate the variance of X_1 and X_3, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.624652, and making a similar calculation for 3 gives Var(X_3)= 2.256604. With these calculations finished, we begin demonstrating the fact that X_1 and X_3 are jointly normally distributed. In order to do so, we calculate E((X[1]-9.719116002)^p1*(X[3]-3.949197980)^p2/(6.624651810^(1/2*p1))/(2.256604355^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[3]*d/dy[3] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_3 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_3 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_3. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_3 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_3 is -0.336013, while the moment would be -0.336013 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_3 is -0.101967, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_3 is -0.984760, while the moment would be -1.008038 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_3 is -0.611675, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_3 is -0.047414, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_3 is 1.203948, while the moment would be 1.225809 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_3 is 0.096626, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_3 is 3.916746, while the moment would be 4.354853 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_3 is -0.992721, while the moment would be -1.008038 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_3 is -0.067697, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_3 is -3.025529, while the moment would be -3.251737 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_3 is -1.246441, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_3 is -0.462771, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_3 is 4.097316, while the moment would be 4.354853 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_3 is -0.254156, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_3 is 14.753982, while the moment would be 17.435057 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(86367275456*n^8+1771089488168*n^7+15173633264284*n^6+70215512147462*n^5+188830918880329*n^4+293557555532912*n^3+242316279227231*n^2+80863041478758*n)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n]+1/8*(-75544840448*n^8-1662476616416*n^7-15597026735488*n^6-81348027774904*n^5-257463090096696*n^4-505107996702144*n^3-598117449257592*n^2-389495949651336*n-106361339814576)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+1]+1/8*(74078699520*n^8+1741330053120*n^7+17631543383424*n^6+100424836970608*n^5+351838674172528*n^4+776146480665380*n^3+1052288869518716*n^2+801178406443872*n+262056499618032)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+2]+1/8*(-39817300992*n^8-995690855040*n^7-10771193872128*n^6-65816150217856*n^5-248327857631632*n^4-592000367818040*n^3-869916296907092*n^2-719387837821524*n-255829062547296)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+3]+1/8*(22223609856*n^8+589069845504*n^7+6751640169216*n^6+43665729800224*n^5+174100330761904*n^4+437602162116296*n^3+675924232731524*n^2+585195750853356*n+216713078305920)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 5, 12, 30, 78, 205, 547, 1488, 4104, 11430, 32111, 90910, 259087, 742701, 2140165, 6195787, 18011450, 52556756, 153881305, 451948784, 1331141032, 3930882046, 11635856286, 34519968211, 102620956036, 305655819202, 912019030930] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_1 and X_5, the numbers of vertices with 1 and 5 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[5])+y[3]*f(x,y[0],y[1],y[3],y[5])^3+y[5]*f(x,y[0],y[1],y[3],y[5])^5). Our next step is to compute E[X_1] and E[X_5]. We do this by finding the total number of vertices with 1 and 5 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[1]*d/dy[1] to multiply every monomial by the number of nodes with 1 children in the corresponding tree, and similarly for 5. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 1 children over all trees with n vertices: b[n+5] = 1/8*(n+4)*(86367275456*n^8+1425620386344*n^7+9860364799780*n^6+36973260867158*n^5+80978768333727*n^4+103292972104564*n^3+70800582548553*n^2+20034229599450*n)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n]+1/8*(n+4)*(-75544840448*n^8-1284752414176*n^7-9249299631616*n^6-36672203100984*n^5-87121905236168*n^4-126090978398000*n^3-107505019194760*n^2-48674745715992*n-8770910726160)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+1]+1/8*(n+4)*(74078699520*n^8+1296857856000*n^7+9665680113024*n^6+39938730293872*n^5+99653168721568*n^4+152834053511524*n^3+139400764659384*n^2+68053831603452*n+13195833777720)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+2]+1/8*(n+4)*(-39817300992*n^8-716969748096*n^7-5513760159744*n^6-23589932709760*n^5-61179358603920*n^4-97946393192456*n^3-93739254528228*n^2-48353658166512*n-10026272320020)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+3]+1/8*(n+4)*(22223609856*n^8+411280966656*n^7+3250412326656*n^6+14281833388576*n^5+37984492396304*n^4+62198256127304*n^3+60588829810444*n^2+31516956717084*n+6458792963040)/(4*n+17)/(2*n+5)/(4*n+15)/(19291328*n^4+125518792*n^3+281835564*n^2+250984878*n+68189913)/(n+3)^2*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [0, 1, 2, 3, 8, 25, 72, 210, 624, 1845, 5470, 16368, 49248, 148590, 449554, 1363650, 4145392, 12625917, 38522970, 117719953, 360229000, 1103691876, 3385388710, 10394822032, 31947384768, 98272051150, 302532263436, 932039141697, 2873386769008, 8864018756858] Doing the same for vertices with 5 children, we obtain the recurrence b[n+5] = 1/8*(12338182208*n^8+254933954264*n^7+2201259221612*n^6+10263955058786*n^5+27799550332697*n^4+43495319251076*n^3+36103285414383*n^2+12102224886774*n)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n]+1/8*(-10792120064*n^8-228384973344*n^7-2040558793152*n^6-10011863043336*n^5-29343484518480*n^4-52232649467196*n^3-54641982059368*n^2-30384216311724*n-6820296304536)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+1]+1/8*(10582671360*n^8+229243914240*n^7+2109453939072*n^6+10723342056624*n^5+32726504852304*n^4+60765882731220*n^3+65797593673788*n^2+36531230915376*n+7151022214416)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+2]+1/8*(-5688185856*n^8-126062696832*n^7-1182937788672*n^6-6089182053120*n^5-18548192368944*n^4-33351086840808*n^3-32551011267108*n^2-12857613995160*n+621975064500)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+3]+1/8*(3174801408*n^8+71947975680*n^7+684275591424*n^6+3515733543648*n^5+10387329102192*n^4+16989198998520*n^3+12177653185956*n^2-1618964503188*n-5114859047640)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+4] and the sequence [0, 0, 0, 0, 0, 1, 6, 21, 64, 198, 622, 1947, 6072, 18889, 58604, 181533, 562048, 1739848, 5384988, 16666154, 51582948, 159670623, 494325128, 1530680360, 4740781896, 14686378600, 45507426250, 141043988475, 437252462220, 1355860399230] Armed with this information, we can compute both the expectation E[X_1] = 9.719116 and E[X_5] = 1.486658. In order to calculate the variance of X_1 and X_5, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[1]*d/dy[1] twice rather than once. Doing so gives Var(X_1)= 6.624652, and making a similar calculation for 5 gives Var(X_5)= 0.765564. With these calculations finished, we begin demonstrating the fact that X_1 and X_5 are jointly normally distributed. In order to do so, we calculate E((X[1]-9.719116002)^p1*(X[5]-1.486658012)^p2/(6.624651810^(1/2*p1))/(.7655637396^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[1]*d/dy[1] p1 times and y[5]*d/dy[5] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_1 and X_5 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_1 and X_5 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_1 and X_5. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_1 and X_5 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_1 and X_5 is -0.242196, while the moment would be -0.242196 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_1 and X_5 is -0.116808, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_1 and X_5 is -0.718205, while the moment would be -0.726589 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_1 and X_5 is -0.812657, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_1 and X_5 is -0.034215, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_1 and X_5 is 1.096931, while the moment would be 1.117318 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_1 and X_5 is 0.235639, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_1 and X_5 is 3.303030, while the moment would be 3.703908 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_1 and X_5 is -0.715586, while the moment would be -0.726589 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_1 and X_5 is -0.155909, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_1 and X_5 is -2.105031, while the moment would be -2.265008 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_1 and X_5 is -1.932640, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_1 and X_5 is -0.333811, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_1 and X_5 is 3.460320, while the moment would be 3.703908 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_1 and X_5 is 0.421520, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_1 and X_5 is 11.032632, while the moment would be 13.306029 if the random variables were normally distributed. This section will consider trees on 30 vertices where the number of children of each vertex lies in {0, 1, 3, 5}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)+x*f(x)^3+x*f(x)^5; the coefficient on the x^n term is the number of trees with n vertices. To extract the relevant coefficient, we can use the Lagrange inversion theorem, which states the the coefficient of x^n in f is the same as the coefficient of z^(n-1) in the expression 1/n*(z^5+z^3+z+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+5] = 1/8*(86367275456*n^8+1771089488168*n^7+15173633264284*n^6+70215512147462*n^5+188830918880329*n^4+293557555532912*n^3+242316279227231*n^2+80863041478758*n)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n]+1/8*(-75544840448*n^8-1662476616416*n^7-15597026735488*n^6-81348027774904*n^5-257463090096696*n^4-505107996702144*n^3-598117449257592*n^2-389495949651336*n-106361339814576)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+1]+1/8*(74078699520*n^8+1741330053120*n^7+17631543383424*n^6+100424836970608*n^5+351838674172528*n^4+776146480665380*n^3+1052288869518716*n^2+801178406443872*n+262056499618032)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+2]+1/8*(-39817300992*n^8-995690855040*n^7-10771193872128*n^6-65816150217856*n^5-248327857631632*n^4-592000367818040*n^3-869916296907092*n^2-719387837821524*n-255829062547296)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+3]+1/8*(22223609856*n^8+589069845504*n^7+6751640169216*n^6+43665729800224*n^5+174100330761904*n^4+437602162116296*n^3+675924232731524*n^2+585195750853356*n+216713078305920)/(4*n+21)/(2*n+7)/(4*n+19)/(n+4)/(19291328*n^4+202684104*n^3+774139908*n^2+1268377694*n+745820475)*b[n+4]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 1, 1, 2, 5, 12, 30, 78, 205, 547, 1488, 4104, 11430, 32111, 90910, 259087, 742701, 2140165, 6195787, 18011450, 52556756, 153881305, 451948784, 1331141032, 3930882046, 11635856286, 34519968211, 102620956036, 305655819202, 912019030930] We are interested in more than just the number of such trees, though. We are investigating the distribution of the statistics X_3 and X_5, the numbers of vertices with 3 and 5 children respectively in a tree chosen uniformly at random. Therefore, we must generalize f to a function of [x, y[0], y[1], y[3], y[5]] which satisfies the similar functional equation f(x,y[0],y[1],y[3],y[5]) = x*(y[0]+y[1]*f(x,y[0],y[1],y[3],y[5])+y[3]*f(x,y[0],y[1],y[3],y[5])^3+y[5]*f(x,y[0],y[1],y[3],y[5])^5). Our next step is to compute E[X_3] and E[X_5]. We do this by finding the total number of vertices with 3 and 5 children over all trees with 30 vertices. Earlier, we used the Lagrange inversion theorem to extract the coefficients of %f, but this time we will first apply the operator y[3]*d/dy[3] to multiply every monomial by the number of nodes with 3 children in the corresponding tree, and similarly for 5. As before, we combine Lagrange inversion with the amazing Almkvist-Zeilberger algorithm to find the following recurrence for the total number of vertices with 3 children over all trees with n vertices: b[n+5] = 1/8*(n+4)*(43183637728*n^8+802539516548*n^7+6250574393294*n^6+26383956008773*n^5+64984279519115*n^4+93009471856125*n^3+71205660425119*n^2+22287730593810*n)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n]+1/8*(n+4)*(-37772420224*n^8-720861806896*n^7-5824942898544*n^6-25937291880060*n^5-69285015591572*n^4-113009667340680*n^3-109072005282372*n^2-56421450343676*n-11869535178600)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+1]+1/8*(n+4)*(37039349760*n^8+725391313920*n^7+6058800922496*n^6+28119836193280*n^5+79047409894388*n^4+137152613281148*n^3+142387742590212*n^2+79890265297748*n+18131421962040)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+2]+1/8*(n+4)*(-19908650496*n^8-399852156480*n^7-3430013629504*n^6-16371160485472*n^5-47381234146192*n^4-84727447649764*n^3-90781694559188*n^2-52738164303084*n-12508434957420)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+3]+1/8*(n+4)*(11111804928*n^8+228729199104*n^7+2008041585536*n^6+9790512953360*n^5+28885142347272*n^4+52544792647716*n^3+57175977965824*n^2+33718519987780*n+8142942662400)/(4*n+17)/(2*n+7)/(4*n+19)/(n+3)/(n+1)/(9645664*n^4+82801684*n^3+255465042*n^2+331738265*n+150061600)*b[n+4]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 0, 1, 4, 10, 26, 77, 232, 696, 2100, 6347, 19184, 58136, 176722, 538330, 1642528, 5019216, 15358752, 47055609, 144330460, 443154138, 1361952966, 4189339968, 12896624952, 39730821000, 122483259600, 377835576300, 1166237952516, 3601743714654] Doing the same for vertices with 5 children, we obtain the recurrence b[n+5] = 1/8*(12338182208*n^8+254933954264*n^7+2201259221612*n^6+10263955058786*n^5+27799550332697*n^4+43495319251076*n^3+36103285414383*n^2+12102224886774*n)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n]+1/8*(-10792120064*n^8-228384973344*n^7-2040558793152*n^6-10011863043336*n^5-29343484518480*n^4-52232649467196*n^3-54641982059368*n^2-30384216311724*n-6820296304536)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+1]+1/8*(10582671360*n^8+229243914240*n^7+2109453939072*n^6+10723342056624*n^5+32726504852304*n^4+60765882731220*n^3+65797593673788*n^2+36531230915376*n+7151022214416)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+2]+1/8*(-5688185856*n^8-126062696832*n^7-1182937788672*n^6-6089182053120*n^5-18548192368944*n^4-33351086840808*n^3-32551011267108*n^2-12857613995160*n+621975064500)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+3]+1/8*(3174801408*n^8+71947975680*n^7+684275591424*n^6+3515733543648*n^5+10387329102192*n^4+16989198998520*n^3+12177653185956*n^2-1618964503188*n-5114859047640)/(n-1)/(4*n+21)/(2*n+7)/(4*n+19)/(2755904*n^4+29383992*n^3+114233764*n^2+190407402*n+113752215)*b[n+4] and the sequence [0, 0, 0, 0, 0, 1, 6, 21, 64, 198, 622, 1947, 6072, 18889, 58604, 181533, 562048, 1739848, 5384988, 16666154, 51582948, 159670623, 494325128, 1530680360, 4740781896, 14686378600, 45507426250, 141043988475, 437252462220, 1355860399230] Armed with this information, we can compute both the expectation E[X_3] = 3.949198 and E[X_5] = 1.486658. In order to calculate the variance of X_3 and X_5, we will need to calculate the second moments of these two variables, rather than just the first moment. Fortunately, this is simply a matter of applying the operator y[3]*d/dy[3] twice rather than once. Doing so gives Var(X_3)= 2.256604, and making a similar calculation for 5 gives Var(X_5)= 0.765564. With these calculations finished, we begin demonstrating the fact that X_3 and X_5 are jointly normally distributed. In order to do so, we calculate E((X[3]-3.949197980)^p1*(X[5]-1.486658012)^p2/(2.256604355^(1/2*p1))/(.7655637396^(1/2*p2))) for all 1 <= p1,p2 <= p. This calculation proceeds much as the previous ones; we begin by applying the operators y[3]*d/dy[3] p1 times and y[5]*d/dy[5] p2 times. We use the Lagrange inversion theorem and Almkvist-Zeilberger algorithm to find a recurrence for the coefficient of x^n, find the 30 th term, and plug in 1 for all y. Since our goal is to determine whether X_3 and X_5 are asymptotically normally distributed, we need to compare these moments to what the moments would be if they were normal. If they were normal, then X_3 and X_5 would have the pdf f(x,y) = 1/2*e^(-1/2*x^2-1/2*y^2+c*x*y)*(-c^2+1)^(1/2)/Pi where c is the correlation between X_3 and X_5. We can compute this correlation, and then compute moments based on this pdf. After computing both the actual moments and what the moments would be if X_3 and X_5 were jointly normal, we find the following results: The (1,1) scaled moment about the mean for X_3 and X_5 is -0.832435, while the moment would be -0.832435 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_3 and X_5 is -0.101028, while the moment would be 0.000000 if the random variables were normally distributed. The (1,3) scaled moment about the mean for X_3 and X_5 is -2.185579, while the moment would be -2.497305 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_3 and X_5 is -1.159686, while the moment would be 0.000000 if the random variables were normally distributed. The (2,1) scaled moment about the mean for X_3 and X_5 is 0.022152, while the moment would be 0.000000 if the random variables were normally distributed. The (2,2) scaled moment about the mean for X_3 and X_5 is 2.081283, while the moment would be 2.385896 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_3 and X_5 is 0.752080, while the moment would be 0.000000 if the random variables were normally distributed. The (2,4) scaled moment about the mean for X_3 and X_5 is 7.633185, while the moment would be 11.315375 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_3 and X_5 is -2.195250, while the moment would be -2.497305 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_3 and X_5 is -0.412889, while the moment would be 0.000000 if the random variables were normally distributed. The (3,3) scaled moment about the mean for X_3 and X_5 is -7.296210, while the moment would be -10.952918 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_3 and X_5 is -5.675999, while the moment would be 0.000000 if the random variables were normally distributed. The (4,1) scaled moment about the mean for X_3 and X_5 is 0.073415, while the moment would be 0.000000 if the random variables were normally distributed. The (4,2) scaled moment about the mean for X_3 and X_5 is 7.594524, while the moment would be 11.315375 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_3 and X_5 is 3.862465, while the moment would be 0.000000 if the random variables were normally distributed. The (4,4) scaled moment about the mean for X_3 and X_5 is 32.092822, while the moment would be 70.416490 if the random variables were normally distributed.