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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.

%I A251576 #14 Jun 04 2025 12:20:55
%S A251576 1,1,6,96,2736,115056,6455376,454666176,38610711936,3842344221696,
%T A251576 438721154343936,56549927146392576,8123473514799876096,
%U A251576 1287034084022760677376,222964032114987212998656,41930788886197036399190016,8507629742037888427727486976,1852490637585980898960109142016
%N 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.
%F A251576 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 A251576 (1) A'(x)/A(x) = G(x)^5.
%F A251576 (2) A'(x) = exp(6*x*G(x)^5).
%F A251576 (3) A(x) = exp( Integral G(x)^5 dx ).
%F A251576 (4) A(x) = exp( Sum_{n>=1} A130564(n)*x^n/n ), where A130564(n) = binomial(6*n-2,n)/(5*n-1).
%F A251576 (5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251586.
%F A251576 (6) A(x) = Sum_{n>=0} A251586(n)*(x/A(x))^n/n! and
%F A251576 (7) [x^n/n!] A(x)^(n+1) = (n+1)*A251586(n),
%F A251576 where A251586(n) = 6^(n-4) * (n+1)^(n-6) * (125*n^4 + 810*n^3 + 2095*n^2 + 2586*n + 1296).
%F A251576 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 A251576 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 A251576 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 A251576 E.g.f.: A(x) = 1 + x + 6*x^2/2! + 96*x^3/3! + 2736*x^4/4! + 115056*x^5/5! +...
%e A251576 such that A(x) = exp(6*x*G(x)^5) / G(x)^5
%e A251576 where G(x) = 1 + x*G(x)^6 is the g.f. of A002295:
%e A251576 G(x) = 1 + x + 6*x^2 + 51*x^3 + 506*x^4 + 5481*x^5 + 62832*x^6 +...
%e A251576 Note that
%e A251576 A'(x) = exp(6*x*G(x)^5) = 1 + 6*x + 96*x^2/2! + 2736*x^3/3! +...
%e A251576 LOGARITHMIC DERIVATIVE.
%e A251576 The logarithm of the e.g.f. begins:
%e A251576 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 A251576 and so A'(x)/A(x) = G(x)^5.
%e A251576 TABLE OF POWERS OF E.G.F.
%e A251576 Form a table of coefficients of x^k/k! in A(x)^n as follows.
%e A251576 n=1: [1, 1,   6,   96,  2736,  115056,   6455376,  454666176, ...];
%e A251576 n=2: [1, 2,  14,  228,  6456,  268992,  14968224, 1047087648, ...];
%e A251576 n=3: [1, 3,  24,  402, 11376,  470808,  26011584, 1808151552, ...];
%e A251576 n=4: [1, 4,  36,  624, 17736,  730944,  40143456, 2774490624, ...];
%e A251576 n=5: [1, 5,  50,  900, 25800, 1061400,  58017600, 3989340000, ...];
%e A251576 n=6: [1, 6,  66, 1236, 35856, 1475856,  80395056, 5503484736, ...];
%e A251576 n=7: [1, 7,  84, 1638, 48216, 1989792, 108156384, 7376303088, ...];
%e A251576 n=8: [1, 8, 104, 2112, 63216, 2620608, 142314624, 9676910592, ...]; ...
%e A251576 in which the main diagonal begins (see A251586):
%e A251576 [1, 2, 24, 624, 25800, 1475856, 108156384, 9676910592, ...]
%e A251576 and is given by the formula:
%e A251576 [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 A251576 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 A251576 (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 A251576 for(n=0, 20, print1(a(n), ", "))
%o A251576 (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 A251576 for(n=0, 20, print1(a(n), ", "))
%Y A251576 Cf. A251586, A251666, A130564, A002295.
%Y A251576 Cf. Variants: A243953, A251573, A251574, A251575, A251577, A251578, A251579, A251580.
%K A251576 nonn
%O A251576 0,3
%A A251576 _Paul D. Hanna_, Dec 06 2014