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A251576 E.g.f.: exp(6*x*G(x)^5) / G(x)^5 where G(x) = 1 + x*G(x)^6 is the g.f. of A002295. 11

%I #11 Dec 09 2014 00:50:43

%S 1,1,6,96,2736,115056,6455376,454666176,38610711936,3842344221696,

%T 438721154343936,56549927146392576,8123473514799876096,

%U 1287034084022760677376,222964032114987212998656,41930788886197036399190016,8507629742037888427727486976,1852490637585980898960109142016

%N E.g.f.: exp(6*x*G(x)^5) / G(x)^5 where G(x) = 1 + x*G(x)^6 is the g.f. of A002295.

%F Let G(x) = 1 + x*G(x)^6 be the g.f. of A002295, then the e.g.f. A(x) of this sequence satisfies:

%F (1) A'(x)/A(x) = G(x)^5.

%F (2) A'(x) = exp(6*x*G(x)^5).

%F (3) A(x) = exp( Integral G(x)^5 dx ).

%F (4) A(x) = exp( Sum_{n>=1} A130564(n)*x^n/n ), where A130564(n) = binomial(6*n-2,n)/(5*n-1).

%F (5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251586.

%F (6) A(x) = Sum_{n>=0} A251586(n)*(x/A(x))^n/n! and

%F (7) [x^n/n!] A(x)^(n+1) = (n+1)*A251586(n),

%F where A251586(n) = 6^(n-4) * (n+1)^(n-6) * (125*n^4 + 810*n^3 + 2095*n^2 + 2586*n + 1296).

%F a(n) = Sum_{k=0..n} 6^k * n!/k! * binomial(6*n-k-6, n-k) * (k-1)/(n-1) for n>1.

%F Recurrence: 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(9*n^4 - 99*n^3 + 413*n^2 - 777*n + 559)*a(n) = 72*(5832*n^9 - 113724*n^8 + 986580*n^7 - 5003586*n^6 + 16373448*n^5 - 35916483*n^4 + 52931854*n^3 - 50678109*n^2 + 28701206*n - 7357350)*a(n-1) + 46656*(9*n^4 - 63*n^3 + 170*n^2 - 212*n + 105)*a(n-2). - _Vaclav Kotesovec_, Dec 07 2014

%F a(n) ~ 6^(6*(n-1)-1/2) / 5^(5*(n-1)-1/2) * n^(n-2) / exp(n-1). - _Vaclav Kotesovec_, Dec 07 2014

%e E.g.f.: A(x) = 1 + x + 6*x^2/2! + 96*x^3/3! + 2736*x^4/4! + 115056*x^5/5! +...

%e such that A(x) = exp(6*x*G(x)^5) / G(x)^5

%e where G(x) = 1 + x*G(x)^6 is the g.f. of A002295:

%e G(x) = 1 + x + 6*x^2 + 51*x^3 + 506*x^4 + 5481*x^5 + 62832*x^6 +...

%e Note that

%e A'(x) = exp(6*x*G(x)^5) = 1 + 6*x + 96*x^2/2! + 2736*x^3/3! +...

%e LOGARITHMIC DERIVATIVE.

%e The logarithm of the e.g.f. begins:

%e log(A(x)) = x + 5*x^2/2 + 40*x^3/3 + 385*x^4/4 + 4095*x^5/5 + 46376*x^6/6 +...

%e and so A'(x)/A(x) = G(x)^5.

%e TABLE OF POWERS OF E.G.F.

%e Form a table of coefficients of x^k/k! in A(x)^n as follows.

%e n=1: [1, 1, 6, 96, 2736, 115056, 6455376, 454666176, ...];

%e n=2: [1, 2, 14, 228, 6456, 268992, 14968224, 1047087648, ...];

%e n=3: [1, 3, 24, 402, 11376, 470808, 26011584, 1808151552, ...];

%e n=4: [1, 4, 36, 624, 17736, 730944, 40143456, 2774490624, ...];

%e n=5: [1, 5, 50, 900, 25800, 1061400, 58017600, 3989340000, ...];

%e n=6: [1, 6, 66, 1236, 35856, 1475856, 80395056, 5503484736, ...];

%e n=7: [1, 7, 84, 1638, 48216, 1989792, 108156384, 7376303088, ...];

%e n=8: [1, 8, 104, 2112, 63216, 2620608, 142314624, 9676910592, ...]; ...

%e in which the main diagonal begins (see A251586):

%e [1, 2, 24, 624, 25800, 1475856, 108156384, 9676910592, ...]

%e and is given by the formula:

%e [x^n/n!] A(x)^(n+1) = 6^(n-4) * (n+1)^(n-5) * (125*n^4 + 810*n^3 + 2095*n^2 + 2586*n + 1296) for n>=0.

%t Flatten[{1,1,Table[Sum[6^k * n!/k! * Binomial[6*n-k-6, n-k] * (k-1)/(n-1),{k,0,n}],{n,2,20}]}] (* _Vaclav Kotesovec_, Dec 07 2014 *)

%o (PARI) {a(n) = local(G=1);for(i=1,n,G=1+x*G^6 +x*O(x^n)); n!*polcoeff(exp(6*x*G^5)/G^5, n)}

%o for(n=0, 20, print1(a(n), ", "))

%o (PARI) {a(n) = if(n==0|n==1, 1, sum(k=0, n, 6^k * n!/k! * binomial(6*n-k-6,n-k) * (k-1)/(n-1) ))}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A251586, A251666, A130564, A002295.

%Y Cf. Variants: A243953, A251573, A251574, A251575, A251577, A251578, A251579, A251580.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Dec 06 2014

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