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A251575 E.g.f.: exp(5*x*G(x)^4) / G(x)^4 where G(x) = 1 + x*G(x)^5 is the g.f. of A002294.

%I A251575 #16 Jun 04 2025 12:33:24
%S A251575 1,1,5,65,1505,51505,2354725,135258625,9373203425,761486105825,
%T A251575 71001537157925,7475144493546625,877222642396170625,
%U A251575 113551974107296500625,16073867927431440597125,2470217878902686107522625,409596824402404827033730625,72890993386914239524503090625,13857243751694786173837746653125
%N A251575 E.g.f.: exp(5*x*G(x)^4) / G(x)^4 where G(x) = 1 + x*G(x)^5 is the g.f. of A002294.
%F A251575 Let G(x) = 1 + x*G(x)^5 be the g.f. of A002294, then the e.g.f. A(x) of this sequence satisfies:
%F A251575 (1) A'(x)/A(x) = G(x)^4.
%F A251575 (2) A'(x) = exp(5*x*G(x)^4).
%F A251575 (3) A(x) = exp( Integral G(x)^4 dx ).
%F A251575 (4) A(x) = exp( Sum_{n>=1} A118971(n-1)*x^n/n ), where A118971(n-1) = binomial(5*n-2,n)/(4*n-1).
%F A251575 (5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251585.
%F A251575 (6) A(x) = Sum_{n>=0} A251585(n)*(x/A(x))^n/n! and
%F A251575 (7) [x^n/n!] A(x)^(n+1) = (n+1)*A251585(n)
%F A251575 where A251585(n) = 5^(n-3) * (n+1)^(n-5) * (16*n^3 + 87*n^2 + 172*n + 125).
%F A251575 a(n) = Sum_{k=0..n} 5^k * n!/k! * binomial(5*n-k-5, n-k) * (k-1)/(n-1) for n>1.
%F A251575 Recurrence: 8*(2*n-3)*(4*n-7)*(4*n-5)*(25*n^3 - 210*n^2 + 598*n - 581)*a(n) = 5*(15625*n^7 - 240625*n^6 + 1592500*n^5 - 5883125*n^4 + 13135350*n^3 - 17781015*n^2 + 13566657*n - 4523904)*a(n-1) - 3125*(25*n^3 - 135*n^2 + 253*n - 168)*a(n-2). - _Vaclav Kotesovec_, Dec 07 2014
%F A251575 a(n) ~ 5^(5*n-11/2) * n^(n-2) / (2^(8*n-9) * exp(n-1)). - _Vaclav Kotesovec_, Dec 07 2014
%e A251575 E.g.f.: A(x) = 1 + x + 5*x^2/2! + 65*x^3/3! + 1505*x^4/4! + 51505*x^5/5! +...
%e A251575 such that A(x) = exp(5*x*G(x)^4) / G(x)^4
%e A251575 where G(x) = 1 + x*G(x)^5 is the g.f. of A002294:
%e A251575 G(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + 23751*x^6 +...
%e A251575 Note that
%e A251575 A'(x) = exp(5*x*G(x)^4) = 1 + 5*x + 65*x^2/2! + 1505*x^3/3! + 51505*x^4/4! +...
%e A251575 LOGARITHMIC DERIVATIVE.
%e A251575 The logarithm of the e.g.f. begins:
%e A251575 log(A(x)) = x + 4*x^2/2 + 26*x^3/3 + 204*x^4/4 + 1771*x^5/5 +...
%e A251575 and so A'(x)/A(x) = G(x)^4.
%e A251575 TABLE OF POWERS OF E.G.F.
%e A251575 Form a table of coefficients of x^k/k! in A(x)^n as follows.
%e A251575 n=1: [1, 1,  5,   65,  1505,   51505,  2354725,  135258625, ...];
%e A251575 n=2: [1, 2, 12,  160,  3680,  124560,  5637760,  321147200, ...];
%e A251575 n=3: [1, 3, 21,  291,  6705,  225315, 10112805,  571694355, ...];
%e A251575 n=4: [1, 4, 32,  464, 10784,  361120, 16101760,  904145920, ...];
%e A251575 n=5: [1, 5, 45,  685, 16145,  540645, 23993725, 1339552925, ...];
%e A251575 n=6: [1, 6, 60,  960, 23040,  774000, 34254720, 1903435200, ...];
%e A251575 n=7: [1, 7, 77, 1295, 31745, 1072855, 47438125, 2626525615, ...];
%e A251575 n=8: [1, 8, 96, 1696, 42560, 1450560, 64195840, 3545600000, ...]; ...
%e A251575 in which the main diagonal begins (see A251585):
%e A251575 [1, 2, 21, 464, 16145, 774000, 47438125, 3545600000, ...]
%e A251575 and is given by the formula:
%e A251575 [x^n/n!] A(x)^(n+1) = 5^(n-3) * (n+1)^(n-4) * (16*n^3 + 87*n^2 + 172*n + 125) for n>=0.
%t A251575 Flatten[{1,1,Table[Sum[5^k * n!/k! * Binomial[5*n-k-5, n-k] * (k-1)/(n-1),{k,0,n}],{n,2,20}]}] (* _Vaclav Kotesovec_, Dec 07 2014 *)
%o A251575 (PARI) {a(n) = local(G=1);for(i=1,n,G=1+x*G^5 +x*O(x^n)); n!*polcoeff(exp(5*x*G^4)/G^4, n)}
%o A251575 for(n=0, 20, print1(a(n), ", "))
%o A251575 (PARI) {a(n) = if(n==0||n==1, 1, sum(k=0, n, 5^k * n!/k! * binomial(5*n-k-5,n-k) * (k-1)/(n-1) ))}
%o A251575 for(n=0, 20, print1(a(n), ", "))
%Y A251575 Cf. A251585, A251665, A002294, A118971.
%Y A251575 Cf. Variants: A243953, A251573, A251574, A251576, A251577, A251578, A251579, A251580.
%K A251575 nonn
%O A251575 0,3
%A A251575 _Paul D. Hanna_, Dec 06 2014