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 50 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 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 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] = 24.680478 and E[X_1] = 16.088084. 

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)= 6.253874, and making a similar calculation for 1 gives Var(X_1)= 10.964354. 

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]-24.68047816)^p1*(X[1]-16.08808407)^p2/(6.253874380^(1/2*p1))/(10.96435352^(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 50 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 -0.955389, while the moment would be -0.955389 if the random variables were normally distributed. 

The (1,2) scaled moment about the mean for X_0 and X_1 is -0.104763, 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 -2.839841, while the moment would be -2.866168 if the random variables were normally distributed. 

The (1,4) scaled moment about the mean for X_0 and X_1 is -1.032289, 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 0.091345, 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 2.799434, while the moment would be 2.825538 if the random variables were normally distributed. 

The (2,3) scaled moment about the mean for X_0 and X_1 is 0.969821, 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 13.675704, while the moment would be 13.953225 if the random variables were normally distributed. 

The (3,1) scaled moment about the mean for X_0 and X_1 is -2.839476, while the moment would be -2.866168 if the random variables were normally distributed. 

The (3,2) scaled moment about the mean for X_0 and X_1 is -0.893757, 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 -13.542020, while the moment would be -13.830801 if the random variables were normally distributed. 

The (3,4) scaled moment about the mean for X_0 and X_1 is -9.556744, 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 0.800534, 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 13.643522, while the moment would be 13.953225 if the random variables were normally distributed. 

The (4,3) scaled moment about the mean for X_0 and X_1 is 8.995391, 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 92.130626, while the moment would be 94.714873 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, 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 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+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] = 24.680478 and E[X_3] = 6.622636. 

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)= 6.253874, and making a similar calculation for 3 gives Var(X_3)= 3.831379. 

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]-24.68047816)^p1*(X[3]-6.622636442)^p2/(6.253874380^(1/2*p1))/(3.831378984^(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 0.038379, while the moment would be 0.038379 if the random variables were normally distributed. 

The (1,2) scaled moment about the mean for X_0 and X_3 is 0.069665, 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 0.156085, while the moment would be 0.115136 if the random variables were normally distributed. 

The (1,4) scaled moment about the mean for X_0 and X_3 is 0.415742, 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 -0.088215, 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 0.980245, while the moment would be 1.002946 if the random variables were normally distributed. 

The (2,3) scaled moment about the mean for X_0 and X_3 is -0.201992, 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 2.747814, while the moment would be 3.017675 if the random variables were normally distributed. 

The (3,1) scaled moment about the mean for X_0 and X_3 is 0.118970, while the moment would be 0.115136 if the random variables were normally distributed. 

The (3,2) scaled moment about the mean for X_0 and X_3 is 0.115018, 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 0.409647, while the moment would be 0.345748 if the random variables were normally distributed. 

The (3,4) scaled moment about the mean for X_0 and X_3 is 0.841363, 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 -0.530936, 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 2.883706, while the moment would be 3.017675 if the random variables were normally distributed. 

The (4,3) scaled moment about the mean for X_0 and X_3 is -1.313956, 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 8.077993, while the moment would be 9.106103 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, 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 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 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] = 24.680478 and E[X_5] = 2.608801. 

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)= 6.253874, and making a similar calculation for 5 gives Var(X_5)= 1.301746. 

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]-24.68047816)^p1*(X[5]-2.608801320)^p2/(6.253874380^(1/2*p1))/(1.301745883^(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 50 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 0.515042, while the moment would be 0.515042 if the random variables were normally distributed. 

The (1,2) scaled moment about the mean for X_0 and X_5 is 0.113430, 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 1.484004, while the moment would be 1.545126 if the random variables were normally distributed. 

The (1,4) scaled moment about the mean for X_0 and X_5 is 0.895162, 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 0.037785, 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 1.501373, while the moment would be 1.530537 if the random variables were normally distributed. 

The (2,3) scaled moment about the mean for X_0 and X_5 is 0.564563, 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 5.682733, while the moment would be 6.183220 if the random variables were normally distributed. 

The (3,1) scaled moment about the mean for X_0 and X_5 is 1.525685, while the moment would be 1.545126 if the random variables were normally distributed. 

The (3,2) scaled moment about the mean for X_0 and X_5 is 0.378924, 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 5.116496, while the moment would be 5.455125 if the random variables were normally distributed. 

The (3,4) scaled moment about the mean for X_0 and X_5 is 4.029364, 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 0.081751, 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 5.880382, while the moment would be 6.183220 if the random variables were normally distributed. 

The (4,3) scaled moment about the mean for X_0 and X_5 is 2.876679, 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 26.437811, while the moment would be 29.788137 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, 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 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[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] = 16.088084 and E[X_3] = 6.622636. 

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)= 10.964354, and making a similar calculation for 3 gives Var(X_3)= 3.831379. 

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]-16.08808407)^p1*(X[3]-6.622636442)^p2/(10.96435352^(1/2*p1))/(3.831378984^(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 50 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 -0.331798, while the moment would be -0.331798 if the random variables were normally distributed. 

The (1,2) scaled moment about the mean for X_1 and X_3 is -0.077147, 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 -0.981782, while the moment would be -0.995395 if the random variables were normally distributed. 

The (1,4) scaled moment about the mean for X_1 and X_3 is -0.491965, 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 -0.036379, 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 1.207350, while the moment would be 1.220180 if the random variables were normally distributed. 

The (2,3) scaled moment about the mean for X_1 and X_3 is 0.081401, 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 4.063662, while the moment would be 4.321082 if the random variables were normally distributed. 

The (3,1) scaled moment about the mean for X_1 and X_3 is -0.986250, while the moment would be -0.995395 if the random variables were normally distributed. 

The (3,2) scaled moment about the mean for X_1 and X_3 is -0.048678, 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 -3.070840, while the moment would be -3.205352 if the random variables were normally distributed. 

The (3,4) scaled moment about the mean for X_1 and X_3 is -1.014756, 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 -0.358473, 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 4.167918, while the moment would be 4.321082 if the random variables were normally distributed. 

The (4,3) scaled moment about the mean for X_1 and X_3 is -0.189825, 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 15.594694, while the moment would be 17.217370 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, 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 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[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] = 16.088084 and E[X_5] = 2.608801. 

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)= 10.964354, and making a similar calculation for 5 gives Var(X_5)= 1.301746. 

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]-16.08808407)^p1*(X[5]-2.608801320)^p2/(10.96435352^(1/2*p1))/(1.301745883^(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 50 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 -0.238903, while the moment would be -0.238903 if the random variables were normally distributed. 

The (1,2) scaled moment about the mean for X_1 and X_5 is -0.088271, 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 -0.712010, while the moment would be -0.716708 if the random variables were normally distributed. 

The (1,4) scaled moment about the mean for X_1 and X_5 is -0.616470, 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 -0.026201, 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 1.102199, while the moment would be 1.114149 if the random variables were normally distributed. 

The (2,3) scaled moment about the mean for X_1 and X_5 is 0.162708, 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 3.453379, while the moment would be 3.684894 if the random variables were normally distributed. 

The (3,1) scaled moment about the mean for X_1 and X_5 is -0.710127, while the moment would be -0.716708 if the random variables were normally distributed. 

The (3,2) scaled moment about the mean for X_1 and X_5 is -0.115922, 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 -2.139792, while the moment would be -2.231935 if the random variables were normally distributed. 

The (3,4) scaled moment about the mean for X_1 and X_5 is -1.461830, 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 -0.258157, 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 3.540063, while the moment would be 3.684894 if the random variables were normally distributed. 

The (4,3) scaled moment about the mean for X_1 and X_5 is 0.262092, 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 11.848460, while the moment would be 13.187542 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, 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 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 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] = 6.622636 and E[X_5] = 2.608801. 

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)= 3.831379, and making a similar calculation for 5 gives Var(X_5)= 1.301746. 

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]-6.622636442)^p1*(X[5]-2.608801320)^p2/(3.831378984^(1/2*p1))/(1.301745883^(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 50 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 -0.836767, while the moment would be -0.836767 if the random variables were normally distributed. 

The (1,2) scaled moment about the mean for X_3 and X_5 is -0.063646, 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 -2.330968, while the moment would be -2.510300 if the random variables were normally distributed. 

The (1,4) scaled moment about the mean for X_3 and X_5 is -0.773440, 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 0.005145, 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 2.224913, while the moment would be 2.400357 if the random variables were normally distributed. 

The (2,3) scaled moment about the mean for X_3 and X_5 is 0.465413, 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 9.137009, while the moment would be 11.402142 if the random variables were normally distributed. 

The (3,1) scaled moment about the mean for X_3 and X_5 is -2.336157, while the moment would be -2.510300 if the random variables were normally distributed. 

The (3,2) scaled moment about the mean for X_3 and X_5 is -0.203801, 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 -8.806113, while the moment would be -11.046216 if the random variables were normally distributed. 

The (3,4) scaled moment about the mean for X_3 and X_5 is -3.869835, 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 -0.060924, 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 9.135452, while the moment would be 11.402142 if the random variables were normally distributed. 

The (4,3) scaled moment about the mean for X_3 and X_5 is 2.304760, 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 45.394754, while the moment would be 71.178848 if the random variables were normally distributed.