On Statistics of Ordered Trees Where the Number of Children a Node Can Have is Restricted to {0, 2, 3, 4}. By Lumpy (AKA Yonah's Computer) This section will consider trees on 50 vertices where the number of children of each vertex lies in {0, 2, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)^2+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^2+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(n+2)*(6007375*n^5+67918675*n^4+271483235*n^3+441844937*n^2+232273002*n)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n]+1/3*(n+2)*(5937250*n^5+76031725*n^4+369426995*n^3+842142239*n^2+890903871*n+348094152)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+1]+1/3*(n+2)*(3155625*n^5+45144000*n^4+252711600*n^3+690548580*n^2+918916851*n+474714216)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+2]+1/3*(n+2)*(-327250*n^5-5172475*n^4-32151095*n^3-98085509*n^2-146591103*n-85669704)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 0, 1, 1, 3, 5, 14, 28, 74, 168, 432, 1045, 2684, 6721, 17355, 44408, 115502, 299812, 785570, 2060094, 5434475, 14362841, 38114760, 101360402, 270373303, 722696570, 1936398635, 5198249550, 13982513625, 37674988080] 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[2], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[2],y[3],y[4]) = x*(y[0]+y[2]*f(x,y[0],y[2],y[3],y[4])^2+y[3]*f(x,y[0],y[2],y[3],y[4])^3+y[4]*f(x,y[0],y[2],y[3],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 50 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*(n+2)*(4805900*n^5+35139610*n^4+91666503*n^3+100281914*n^2+38949121*n)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n]+1/3*(n+2)*(4749800*n^5+37104320*n^4+107197886*n^3+139887112*n^2+79601498*n+14557752)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+1]+1/3*(n+2)*(2524500*n^5+20983050*n^4+65306275*n^3+92943556*n^2+57765837*n+10896006)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+2]+1/3*(n+2)*(-261800*n^5-2306920*n^4-7797296*n^3-12509598*n^2-9419394*n-2654424)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [1, 0, 2, 3, 10, 20, 65, 147, 434, 1092, 3072, 8085, 22429, 60346, 166595, 453908, 1252290, 3436992, 9493616, 26176566, 72429520, 200350440, 555353603, 1539864766, 4275507053, 11877336620, 33027475600, 91892232120, 255864996495, 712823093430] Doing the same for vertices with 2 children, we obtain the recurrence b[n+4] = 1/3*(24029500*n^6+271533350*n^5+1185940635*n^4+2490552562*n^3+2498158991*n^2+946043214*n)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n]+1/3*(23749000*n^6+280238200*n^5+1313487670*n^4+3109397344*n^3+3891274030*n^2+2419158532*n+580283376)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+1]+1/3*(12622500*n^6+155256750*n^5+770143275*n^4+1965841080*n^3+2711423721*n^2+1901707362*n+522532080)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+2]+1/3*(-1309000*n^6-16755200*n^5-82920420*n^4-197601574*n^3-225335024*n^2-97324470*n)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+3] and the sequence [0, 0, 1, 0, 4, 5, 21, 42, 140, 324, 975, 2475, 7062, 18733, 52416, 141960, 394368, 1080044, 2994210, 8252004, 22881320, 63305739, 175715078, 487434607, 1354666788, 3765183950, 10477426440, 29165932575, 81256746690, 226481143980] Armed with this information, we can compute both the expectation E[X_0] = 31.348471 and E[X_2] = 10.017773. 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)= 1.351533, and making a similar calculation for 2 gives Var(X_2)= 6.891093. 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]-31.34847113)^p1*(X[2]-10.01777254)^p2/(1.351532934^(1/2*p1))/(6.891092779^(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 50 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 -0.934432, while the moment would be -0.934432 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_2 is -0.116718, 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 -2.741739, while the moment would be -2.803295 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_2 is -1.134269, 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 0.082519, 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 2.683695, while the moment would be 2.746326 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_2 is 0.988316, 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 12.705320, while the moment would be 13.477954 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_2 is -2.737126, while the moment would be -2.803295 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_2 is -0.803787, 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 -12.493346, while the moment would be -13.305353 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_2 is -9.070775, 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 0.565578, 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 12.547322, while the moment would be 13.477954 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_2 is 7.717116, 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 76.975442, while the moment would be 90.165644 if the random variables were normally distributed. This section will consider trees on 50 vertices where the number of children of each vertex lies in {0, 2, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)^2+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^2+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(n+2)*(6007375*n^5+67918675*n^4+271483235*n^3+441844937*n^2+232273002*n)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n]+1/3*(n+2)*(5937250*n^5+76031725*n^4+369426995*n^3+842142239*n^2+890903871*n+348094152)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+1]+1/3*(n+2)*(3155625*n^5+45144000*n^4+252711600*n^3+690548580*n^2+918916851*n+474714216)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+2]+1/3*(n+2)*(-327250*n^5-5172475*n^4-32151095*n^3-98085509*n^2-146591103*n-85669704)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 0, 1, 1, 3, 5, 14, 28, 74, 168, 432, 1045, 2684, 6721, 17355, 44408, 115502, 299812, 785570, 2060094, 5434475, 14362841, 38114760, 101360402, 270373303, 722696570, 1936398635, 5198249550, 13982513625, 37674988080] 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[2], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[2],y[3],y[4]) = x*(y[0]+y[2]*f(x,y[0],y[2],y[3],y[4])^2+y[3]*f(x,y[0],y[2],y[3],y[4])^3+y[4]*f(x,y[0],y[2],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 50 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+4] = 1/3*(n+2)*(4805900*n^5+35139610*n^4+91666503*n^3+100281914*n^2+38949121*n)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n]+1/3*(n+2)*(4749800*n^5+37104320*n^4+107197886*n^3+139887112*n^2+79601498*n+14557752)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+1]+1/3*(n+2)*(2524500*n^5+20983050*n^4+65306275*n^3+92943556*n^2+57765837*n+10896006)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+2]+1/3*(n+2)*(-261800*n^5-2306920*n^4-7797296*n^3-12509598*n^2-9419394*n-2654424)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [1, 0, 2, 3, 10, 20, 65, 147, 434, 1092, 3072, 8085, 22429, 60346, 166595, 453908, 1252290, 3436992, 9493616, 26176566, 72429520, 200350440, 555353603, 1539864766, 4275507053, 11877336620, 33027475600, 91892232120, 255864996495, 712823093430] Doing the same for vertices with 3 children, we obtain the recurrence b[n+4] = 1/3*(n+3)*(1201475*n^6+14771075*n^5+69905285*n^4+158140067*n^3+169403606*n^2+67599224*n)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n]+1/3*(n+3)*(1187450*n^6+15192375*n^5+77024670*n^4+196816529*n^3+265578820*n^2+178236532*n+46454496)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+1]+1/3*(n+3)*(631125*n^6+8390250*n^5+44885425*n^4+123249780*n^3+182284316*n^2+136400688*n+39591552)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+2]+1/3*(n+3)*(-65450*n^6-902825*n^5-4578895*n^4-9813259*n^3-5121639*n^2+10037172*n+10444896)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+3] and the sequence [0, 0, 0, 1, 0, 5, 6, 28, 56, 192, 450, 1375, 3520, 10153, 27118, 76440, 208208, 581604, 1600176, 4455462, 12325680, 34297109, 95188632, 264981068, 737009240, 2053314650, 5719932790, 15950418120, 44487719640, 124169253600] Armed with this information, we can compute both the expectation E[X_0] = 31.348471 and E[X_3] = 5.570570. 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)= 1.351533, and making a similar calculation for 3 gives Var(X_3)= 3.561587. 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]-31.34847113)^p1*(X[3]-5.570570433)^p2/(1.351532934^(1/2*p1))/(3.561587160^(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 50 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 0.135498, while the moment would be 0.135498 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_3 is 0.058795, 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 0.417218, while the moment would be 0.406495 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_3 is 0.417357, 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 -0.160898, 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 1.010489, while the moment would be 1.036720 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_3 is -0.272624, 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 2.904225, while the moment would be 3.220317 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_3 is 0.404687, while the moment would be 0.406495 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_3 is 0.010454, 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 1.145819, while the moment would be 1.234410 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_3 is 0.359241, 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 -0.947418, 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 3.140305, while the moment would be 3.220317 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_3 is -2.288695, 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 8.640203, while the moment would be 10.329992 if the random variables were normally distributed. This section will consider trees on 50 vertices where the number of children of each vertex lies in {0, 2, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)^2+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^2+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(n+2)*(6007375*n^5+67918675*n^4+271483235*n^3+441844937*n^2+232273002*n)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n]+1/3*(n+2)*(5937250*n^5+76031725*n^4+369426995*n^3+842142239*n^2+890903871*n+348094152)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+1]+1/3*(n+2)*(3155625*n^5+45144000*n^4+252711600*n^3+690548580*n^2+918916851*n+474714216)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+2]+1/3*(n+2)*(-327250*n^5-5172475*n^4-32151095*n^3-98085509*n^2-146591103*n-85669704)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 0, 1, 1, 3, 5, 14, 28, 74, 168, 432, 1045, 2684, 6721, 17355, 44408, 115502, 299812, 785570, 2060094, 5434475, 14362841, 38114760, 101360402, 270373303, 722696570, 1936398635, 5198249550, 13982513625, 37674988080] 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[2], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[2],y[3],y[4]) = x*(y[0]+y[2]*f(x,y[0],y[2],y[3],y[4])^2+y[3]*f(x,y[0],y[2],y[3],y[4])^3+y[4]*f(x,y[0],y[2],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 50 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*(n+2)*(4805900*n^5+35139610*n^4+91666503*n^3+100281914*n^2+38949121*n)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n]+1/3*(n+2)*(4749800*n^5+37104320*n^4+107197886*n^3+139887112*n^2+79601498*n+14557752)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+1]+1/3*(n+2)*(2524500*n^5+20983050*n^4+65306275*n^3+92943556*n^2+57765837*n+10896006)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+2]+1/3*(n+2)*(-261800*n^5-2306920*n^4-7797296*n^3-12509598*n^2-9419394*n-2654424)/(3*n+8)/(3*n+7)/(n+3)/(18700*n^3+61930*n^2+58689*n+12234)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [1, 0, 2, 3, 10, 20, 65, 147, 434, 1092, 3072, 8085, 22429, 60346, 166595, 453908, 1252290, 3436992, 9493616, 26176566, 72429520, 200350440, 555353603, 1539864766, 4275507053, 11877336620, 33027475600, 91892232120, 255864996495, 712823093430] Doing the same for vertices with 4 children, we obtain the recurrence b[n+4] = 1/3*(n+3)*(4805900*n^6+63861930*n^5+325262027*n^4+786595788*n^3+891899225*n^2+371509434*n)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n]+1/3*(n+3)*(4749800*n^6+65491360*n^5+356670254*n^4+975145852*n^3+1402251698*n^2+999793180*n+276758640)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+1]+1/3*(n+3)*(2524500*n^6+36070650*n^5+206478015*n^4+602567814*n^3+936858117*n^2+722417748*n+206990100)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+2]+1/3*(n+3)*(-261800*n^6-3871560*n^5-19810424*n^4-32996526*n^3+42794758*n^2+194349960*n+161705160)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+3] and the sequence [0, 0, 0, 0, 1, 0, 6, 7, 36, 72, 255, 605, 1881, 4862, 14196, 38220, 108668, 297976, 837828, 2317848, 6487455, 18029214, 50382167, 140369207, 392149494, 1094275600, 3057928315, 8542404585, 23883432300, 66776151390] Armed with this information, we can compute both the expectation E[X_0] = 31.348471 and E[X_4] = 3.063186. 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)= 1.351533, and making a similar calculation for 4 gives Var(X_4)= 1.944648. 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]-31.34847113)^p1*(X[4]-3.063185909)^p2/(1.351532934^(1/2*p1))/(1.944647596^(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 50 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 0.741981, while the moment would be 0.741981 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_0 and X_4 is 0.189082, 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 2.152589, while the moment would be 2.225943 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_0 and X_4 is 1.742735, 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 0.085640, 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 2.039973, while the moment would be 2.101072 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_0 and X_4 is 1.276265, 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 9.007899, while the moment would be 9.606432 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_0 and X_4 is 2.165726, while the moment would be 2.225943 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_0 and X_4 is 0.887845, 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 8.480654, while the moment would be 9.128755 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_0 and X_4 is 10.189590, 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 0.413385, 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 8.856555, while the moment would be 9.606432 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_0 and X_4 is 7.848956, 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 49.839880, while the moment would be 55.912752 if the random variables were normally distributed. This section will consider trees on 50 vertices where the number of children of each vertex lies in {0, 2, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)^2+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^2+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(n+2)*(6007375*n^5+67918675*n^4+271483235*n^3+441844937*n^2+232273002*n)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n]+1/3*(n+2)*(5937250*n^5+76031725*n^4+369426995*n^3+842142239*n^2+890903871*n+348094152)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+1]+1/3*(n+2)*(3155625*n^5+45144000*n^4+252711600*n^3+690548580*n^2+918916851*n+474714216)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+2]+1/3*(n+2)*(-327250*n^5-5172475*n^4-32151095*n^3-98085509*n^2-146591103*n-85669704)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 0, 1, 1, 3, 5, 14, 28, 74, 168, 432, 1045, 2684, 6721, 17355, 44408, 115502, 299812, 785570, 2060094, 5434475, 14362841, 38114760, 101360402, 270373303, 722696570, 1936398635, 5198249550, 13982513625, 37674988080] 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[2], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[2],y[3],y[4]) = x*(y[0]+y[2]*f(x,y[0],y[2],y[3],y[4])^2+y[3]*f(x,y[0],y[2],y[3],y[4])^3+y[4]*f(x,y[0],y[2],y[3],y[4])^4). 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 50 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+4] = 1/3*(24029500*n^6+271533350*n^5+1185940635*n^4+2490552562*n^3+2498158991*n^2+946043214*n)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n]+1/3*(23749000*n^6+280238200*n^5+1313487670*n^4+3109397344*n^3+3891274030*n^2+2419158532*n+580283376)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+1]+1/3*(12622500*n^6+155256750*n^5+770143275*n^4+1965841080*n^3+2711423721*n^2+1901707362*n+522532080)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+2]+1/3*(-1309000*n^6-16755200*n^5-82920420*n^4-197601574*n^3-225335024*n^2-97324470*n)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 1, 0, 4, 5, 21, 42, 140, 324, 975, 2475, 7062, 18733, 52416, 141960, 394368, 1080044, 2994210, 8252004, 22881320, 63305739, 175715078, 487434607, 1354666788, 3765183950, 10477426440, 29165932575, 81256746690, 226481143980] Doing the same for vertices with 3 children, we obtain the recurrence b[n+4] = 1/3*(n+3)*(1201475*n^6+14771075*n^5+69905285*n^4+158140067*n^3+169403606*n^2+67599224*n)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n]+1/3*(n+3)*(1187450*n^6+15192375*n^5+77024670*n^4+196816529*n^3+265578820*n^2+178236532*n+46454496)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+1]+1/3*(n+3)*(631125*n^6+8390250*n^5+44885425*n^4+123249780*n^3+182284316*n^2+136400688*n+39591552)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+2]+1/3*(n+3)*(-65450*n^6-902825*n^5-4578895*n^4-9813259*n^3-5121639*n^2+10037172*n+10444896)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+3] and the sequence [0, 0, 0, 1, 0, 5, 6, 28, 56, 192, 450, 1375, 3520, 10153, 27118, 76440, 208208, 581604, 1600176, 4455462, 12325680, 34297109, 95188632, 264981068, 737009240, 2053314650, 5719932790, 15950418120, 44487719640, 124169253600] Armed with this information, we can compute both the expectation E[X_2] = 10.017773 and E[X_3] = 5.570570. 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)= 6.891093, and making a similar calculation for 3 gives Var(X_3)= 3.561587. 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]-10.01777254)^p1*(X[3]-5.570570433)^p2/(6.891092779^(1/2*p1))/(3.561587160^(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 50 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 -0.479472, while the moment would be -0.479472 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_2 and X_3 is -0.104189, 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 -1.408134, while the moment would be -1.438416 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_2 and X_3 is -0.865264, 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 -0.070055, 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 1.431721, while the moment would be 1.459787 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_2 and X_3 is 0.230009, 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 5.393475, while the moment would be 5.758720 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_2 and X_3 is -1.413351, while the moment would be -1.438416 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_2 and X_3 is 0.015892, 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 -4.612193, while the moment would be -4.976612 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_2 and X_3 is -1.824587, 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 -0.643923, 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 5.411416, while the moment would be 5.758720 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_2 and X_3 is -0.286712, 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 23.088043, while the moment would be 26.820745 if the random variables were normally distributed. This section will consider trees on 50 vertices where the number of children of each vertex lies in {0, 2, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)^2+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^2+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(n+2)*(6007375*n^5+67918675*n^4+271483235*n^3+441844937*n^2+232273002*n)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n]+1/3*(n+2)*(5937250*n^5+76031725*n^4+369426995*n^3+842142239*n^2+890903871*n+348094152)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+1]+1/3*(n+2)*(3155625*n^5+45144000*n^4+252711600*n^3+690548580*n^2+918916851*n+474714216)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+2]+1/3*(n+2)*(-327250*n^5-5172475*n^4-32151095*n^3-98085509*n^2-146591103*n-85669704)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 0, 1, 1, 3, 5, 14, 28, 74, 168, 432, 1045, 2684, 6721, 17355, 44408, 115502, 299812, 785570, 2060094, 5434475, 14362841, 38114760, 101360402, 270373303, 722696570, 1936398635, 5198249550, 13982513625, 37674988080] 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[2], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[2],y[3],y[4]) = x*(y[0]+y[2]*f(x,y[0],y[2],y[3],y[4])^2+y[3]*f(x,y[0],y[2],y[3],y[4])^3+y[4]*f(x,y[0],y[2],y[3],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 50 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*(24029500*n^6+271533350*n^5+1185940635*n^4+2490552562*n^3+2498158991*n^2+946043214*n)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n]+1/3*(23749000*n^6+280238200*n^5+1313487670*n^4+3109397344*n^3+3891274030*n^2+2419158532*n+580283376)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+1]+1/3*(12622500*n^6+155256750*n^5+770143275*n^4+1965841080*n^3+2711423721*n^2+1901707362*n+522532080)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+2]+1/3*(-1309000*n^6-16755200*n^5-82920420*n^4-197601574*n^3-225335024*n^2-97324470*n)/(3*n+11)/(3*n+10)/(93500*n^3+495550*n^2+827805*n+423696)/(n+1)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 1, 0, 4, 5, 21, 42, 140, 324, 975, 2475, 7062, 18733, 52416, 141960, 394368, 1080044, 2994210, 8252004, 22881320, 63305739, 175715078, 487434607, 1354666788, 3765183950, 10477426440, 29165932575, 81256746690, 226481143980] Doing the same for vertices with 4 children, we obtain the recurrence b[n+4] = 1/3*(n+3)*(4805900*n^6+63861930*n^5+325262027*n^4+786595788*n^3+891899225*n^2+371509434*n)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n]+1/3*(n+3)*(4749800*n^6+65491360*n^5+356670254*n^4+975145852*n^3+1402251698*n^2+999793180*n+276758640)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+1]+1/3*(n+3)*(2524500*n^6+36070650*n^5+206478015*n^4+602567814*n^3+936858117*n^2+722417748*n+206990100)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+2]+1/3*(n+3)*(-261800*n^6-3871560*n^5-19810424*n^4-32996526*n^3+42794758*n^2+194349960*n+161705160)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+3] and the sequence [0, 0, 0, 0, 1, 0, 6, 7, 36, 72, 255, 605, 1881, 4862, 14196, 38220, 108668, 297976, 837828, 2317848, 6487455, 18029214, 50382167, 140369207, 392149494, 1094275600, 3057928315, 8542404585, 23883432300, 66776151390] Armed with this information, we can compute both the expectation E[X_2] = 10.017773 and E[X_4] = 3.063186. 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)= 6.891093, and making a similar calculation for 4 gives Var(X_4)= 1.944648. 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]-10.01777254)^p1*(X[4]-3.063185909)^p2/(6.891092779^(1/2*p1))/(1.944647596^(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 50 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 -0.454565, while the moment would be -0.454565 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_2 and X_4 is -0.127295, 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 -1.328354, while the moment would be -1.363695 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_2 and X_4 is -1.135579, 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 -0.049901, 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 1.375951, while the moment would be 1.413259 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_2 and X_4 is 0.419571, 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 5.074167, while the moment would be 5.479554 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_2 and X_4 is -1.329337, while the moment would be -1.363695 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_2 and X_4 is -0.118882, 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 -4.219034, while the moment would be -4.654646 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_2 and X_4 is -3.022731, 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 -0.509879, 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 5.025863, while the moment would be 5.479554 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_2 and X_4 is 0.804277, 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 20.649608, while the moment would be 24.902020 if the random variables were normally distributed. This section will consider trees on 50 vertices where the number of children of each vertex lies in {0, 2, 3, 4}. These trees have a generating function given by the functional equation f(x) = x+x*f(x)^2+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^2+1)^n. Finally, we can use the amazing Almkvist-Zeilberger algorithm to find that these coefficients follow the recurrence: b[n+4] = 1/3*(n+2)*(6007375*n^5+67918675*n^4+271483235*n^3+441844937*n^2+232273002*n)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n]+1/3*(n+2)*(5937250*n^5+76031725*n^4+369426995*n^3+842142239*n^2+890903871*n+348094152)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+1]+1/3*(n+2)*(3155625*n^5+45144000*n^4+252711600*n^3+690548580*n^2+918916851*n+474714216)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+2]+1/3*(n+2)*(-327250*n^5-5172475*n^4-32151095*n^3-98085509*n^2-146591103*n-85669704)/(3*n+11)/(3*n+13)/(n+4)/(23375*n^3+170775*n^2+403780*n+305856)*b[n+3]. Using this recurrence we can generate the first 30 terms of the enumeration sequence: [1, 0, 1, 1, 3, 5, 14, 28, 74, 168, 432, 1045, 2684, 6721, 17355, 44408, 115502, 299812, 785570, 2060094, 5434475, 14362841, 38114760, 101360402, 270373303, 722696570, 1936398635, 5198249550, 13982513625, 37674988080] 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[2], y[3], y[4]] which satisfies the similar functional equation f(x,y[0],y[2],y[3],y[4]) = x*(y[0]+y[2]*f(x,y[0],y[2],y[3],y[4])^2+y[3]*f(x,y[0],y[2],y[3],y[4])^3+y[4]*f(x,y[0],y[2],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 50 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+4] = 1/3*(n+3)*(1201475*n^6+14771075*n^5+69905285*n^4+158140067*n^3+169403606*n^2+67599224*n)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n]+1/3*(n+3)*(1187450*n^6+15192375*n^5+77024670*n^4+196816529*n^3+265578820*n^2+178236532*n+46454496)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+1]+1/3*(n+3)*(631125*n^6+8390250*n^5+44885425*n^4+123249780*n^3+182284316*n^2+136400688*n+39591552)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+2]+1/3*(n+3)*(-65450*n^6-902825*n^5-4578895*n^4-9813259*n^3-5121639*n^2+10037172*n+10444896)/n/(3*n+11)/(3*n+10)/(n+4)/(4675*n^3+29425*n^2+59430*n+37986)*b[n+3]. Then, this recurrence gives the following sequence of vertex counts: [0, 0, 0, 1, 0, 5, 6, 28, 56, 192, 450, 1375, 3520, 10153, 27118, 76440, 208208, 581604, 1600176, 4455462, 12325680, 34297109, 95188632, 264981068, 737009240, 2053314650, 5719932790, 15950418120, 44487719640, 124169253600] Doing the same for vertices with 4 children, we obtain the recurrence b[n+4] = 1/3*(n+3)*(4805900*n^6+63861930*n^5+325262027*n^4+786595788*n^3+891899225*n^2+371509434*n)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n]+1/3*(n+3)*(4749800*n^6+65491360*n^5+356670254*n^4+975145852*n^3+1402251698*n^2+999793180*n+276758640)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+1]+1/3*(n+3)*(2524500*n^6+36070650*n^5+206478015*n^4+602567814*n^3+936858117*n^2+722417748*n+206990100)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+2]+1/3*(n+3)*(-261800*n^6-3871560*n^5-19810424*n^4-32996526*n^3+42794758*n^2+194349960*n+161705160)/(n-1)/(3*n+11)/(3*n+13)/(n+4)/(18700*n^3+136290*n^2+322361*n+245430)*b[n+3] and the sequence [0, 0, 0, 0, 1, 0, 6, 7, 36, 72, 255, 605, 1881, 4862, 14196, 38220, 108668, 297976, 837828, 2317848, 6487455, 18029214, 50382167, 140369207, 392149494, 1094275600, 3057928315, 8542404585, 23883432300, 66776151390] Armed with this information, we can compute both the expectation E[X_3] = 5.570570 and E[X_4] = 3.063186. 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)= 3.561587, and making a similar calculation for 4 gives Var(X_4)= 1.944648. 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]-5.570570433)^p1*(X[4]-3.063185909)^p2/(3.561587160^(1/2*p1))/(1.944647596^(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 50 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 -0.563701, while the moment would be -0.563701 if the random variables were normally distributed. The (1,2) scaled moment about the mean for X_3 and X_4 is -0.111777, 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 -1.608670, while the moment would be -1.691102 if the random variables were normally distributed. The (1,4) scaled moment about the mean for X_3 and X_4 is -1.135049, 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 -0.049085, 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 1.554510, while the moment would be 1.635517 if the random variables were normally distributed. The (2,3) scaled moment about the mean for X_3 and X_4 is 0.441590, 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 5.917426, while the moment would be 6.813103 if the random variables were normally distributed. The (3,1) scaled moment about the mean for X_3 and X_4 is -1.607286, while the moment would be -1.691102 if the random variables were normally distributed. The (3,2) scaled moment about the mean for X_3 and X_4 is -0.029040, 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 -5.203946, while the moment would be -6.148032 if the random variables were normally distributed. The (3,4) scaled moment about the mean for X_3 and X_4 is -2.959657, 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 -0.585039, 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 5.832836, while the moment would be 6.813103 if the random variables were normally distributed. The (4,3) scaled moment about the mean for X_3 and X_4 is 0.512310, 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 24.734500, while the moment would be 34.301913 if the random variables were normally distributed.