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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. 11

%I

%S 1,1,5,65,1505,51505,2354725,135258625,9373203425,761486105825,

%T 71001537157925,7475144493546625,877222642396170625,

%U 113551974107296500625,16073867927431440597125,2470217878902686107522625,409596824402404827033730625,72890993386914239524503090625,13857243751694786173837746653125

%N 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 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 (1) A'(x)/A(x) = G(x)^4.

%F (2) A'(x) = exp(5*x*G(x)^4).

%F (3) A(x) = exp( Integral G(x)^4 dx ).

%F (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 (5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251585.

%F (6) A(x) = Sum_{n>=0} A251585(n)*(x/A(x))^n/n! and

%F (7) [x^n/n!] A(x)^(n+1) = (n+1)*A251585(n)

%F where A251585(n) = 5^(n-3) * (n+1)^(n-5) * (16*n^3 + 87*n^2 + 172*n + 125).

%F 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 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 a(n) ~ 5^(5*n-11/2) * n^(n-2) / (2^(8*n-9) * exp(n-1)). - _Vaclav Kotesovec_, Dec 07 2014

%e E.g.f.: A(x) = 1 + x + 5*x^2/2! + 65*x^3/3! + 1505*x^4/4! + 51505*x^5/5! +...

%e such that A(x) = exp(5*x*G(x)^4) / G(x)^4

%e where G(x) = 1 + x*G(x)^5 is the g.f. of A002294:

%e G(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + 23751*x^6 +...

%e Note that

%e A'(x) = exp(5*x*G(x)^4) = 1 + 5*x + 65*x^2/2! + 1505*x^3/3! + 51505*x^4/4! +...

%e LOGARITHMIC DERIVATIVE.

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

%e log(A(x)) = x + 4*x^2/2 + 26*x^3/3 + 204*x^4/4 + 1771*x^5/5 +...

%e and so A'(x)/A(x) = G(x)^4.

%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, 5, 65, 1505, 51505, 2354725, 135258625, ...];

%e n=2: [1, 2, 12, 160, 3680, 124560, 5637760, 321147200, ...];

%e n=3: [1, 3, 21, 291, 6705, 225315, 10112805, 571694355, ...];

%e n=4: [1, 4, 32, 464, 10784, 361120, 16101760, 904145920, ...];

%e n=5: [1, 5, 45, 685, 16145, 540645, 23993725, 1339552925, ...];

%e n=6: [1, 6, 60, 960, 23040, 774000, 34254720, 1903435200, ...];

%e n=7: [1, 7, 77, 1295, 31745, 1072855, 47438125, 2626525615, ...];

%e n=8: [1, 8, 96, 1696, 42560, 1450560, 64195840, 3545600000, ...]; ...

%e in which the main diagonal begins (see A251585):

%e [1, 2, 21, 464, 16145, 774000, 47438125, 3545600000, ...]

%e and is given by the formula:

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

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

%Y Cf. A251585, A251665, A002294, A118971.

%Y Cf. Variants: A243953, A251573, A251574, A251576, A251577, A251578, A251579, A251580.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Dec 06 2014

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Last modified April 19 02:47 EDT 2019. Contains 322237 sequences. (Running on oeis4.)