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.

1, 1, 5, 65, 1505, 51505, 2354725, 135258625, 9373203425, 761486105825, 71001537157925, 7475144493546625, 877222642396170625, 113551974107296500625, 16073867927431440597125, 2470217878902686107522625, 409596824402404827033730625, 72890993386914239524503090625, 13857243751694786173837746653125

Offset

0,3

Links

Table of n, a(n) for n=0..18.

Formula

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:

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

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

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

(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).

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

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

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

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

a(n) = Sum_{k=0..n} 5^k * n!/k! * binomial(5*n-k-5, n-k) * (k-1)/(n-1) for n>1.

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

a(n) ~ 5^(5*n-11/2) * n^(n-2) / (2^(8*n-9) * exp(n-1)). - Vaclav Kotesovec, Dec 07 2014

Example

E.g.f.: A(x) = 1 + x + 5*x^2/2! + 65*x^3/3! + 1505*x^4/4! + 51505*x^5/5! +...
such that A(x) = exp(5*x*G(x)^4) / G(x)^4
where G(x) = 1 + x*G(x)^5 is the g.f. of A002294:
G(x) = 1 + x + 5*x^2 + 35*x^3 + 285*x^4 + 2530*x^5 + 23751*x^6 +...
Note that
A'(x) = exp(5*x*G(x)^4) = 1 + 5*x + 65*x^2/2! + 1505*x^3/3! + 51505*x^4/4! +...
LOGARITHMIC DERIVATIVE.
The logarithm of the e.g.f. begins:
log(A(x)) = x + 4*x^2/2 + 26*x^3/3 + 204*x^4/4 + 1771*x^5/5 +...
and so A'(x)/A(x) = G(x)^4.
TABLE OF POWERS OF E.G.F.
Form a table of coefficients of x^k/k! in A(x)^n as follows.
n=1: [1, 1,  5,   65,  1505,   51505,  2354725,  135258625, ...];
n=2: [1, 2, 12,  160,  3680,  124560,  5637760,  321147200, ...];
n=3: [1, 3, 21,  291,  6705,  225315, 10112805,  571694355, ...];
n=4: [1, 4, 32,  464, 10784,  361120, 16101760,  904145920, ...];
n=5: [1, 5, 45,  685, 16145,  540645, 23993725, 1339552925, ...];
n=6: [1, 6, 60,  960, 23040,  774000, 34254720, 1903435200, ...];
n=7: [1, 7, 77, 1295, 31745, 1072855, 47438125, 2626525615, ...];
n=8: [1, 8, 96, 1696, 42560, 1450560, 64195840, 3545600000, ...]; ...
in which the main diagonal begins (see A251585):
[1, 2, 21, 464, 16145, 774000, 47438125, 3545600000, ...]
and is given by the formula:
[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.

Mathematica

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 *)

Prog

(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)}
for(n=0, 20, print1(a(n), ", "))
(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) ))}
for(n=0, 20, print1(a(n), ", "))

Crossrefs

Cf. A251585, A251665, A002294, A118971.

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

Sequence in context: A349470 A234295 A370261 * A277347 A276755 A346115

Adjacent sequences: A251572 A251573 A251574 * A251576 A251577 A251578

Keyword

nonn

Author

Paul D. Hanna, Dec 06 2014

Status

approved