A316702 E.g.f.: Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x^2.

1, 1, 2, 12, 84, 640, 6060, 70728, 941808, 13950144, 230971680, 4242680640, 85192002720, 1854377366400, 43570277097984, 1099505252240640, 29642211339068160, 850166713775554560, 25852506567901839360, 830856828456304128000, 28137892587325700198400, 1001532282143426144133120, 37379628178079964459217920, 1459734364264707546159513600

Offset: 0

Author

Paul D. Hanna, Jul 13 2018

Keywords

• nonn

Comments

More generally, we have the following identity. Given the biexponential series

W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,

then for fixed p and q,

Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k + p)*x + (k + q)*y = W(x,y)^(p+q+1) / ( (1 + x*W(x,y))^q * (1 + y*W(x,y))^p ).

Further, W(x,y) satisfies the biexponential functional equation

( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.

Examples

			E.g.f.: A(x) = 1 + x + 2*x^2/2! + 12*x^3/3! + 84*x^4/4! + 640*x^5/5! + 6060*x^6/6! + 70728*x^7/7! + 941808*x^8/8! + 13950144*x^9/9! + 230971680*x^10/10! + ...
such that
A(x) = 1 + (1 + x^2)*x + (2 + x^2)*(1 + 2*x^2)*x^2/2! + (3 + x^2)*(2 + 2*x^2)*(1 + 3*x^2)*x^3/3! + (4 + x^2)*(3 + 2*x^2)*(2 + 3*x^2)*(1 + 4*x^2)*x^4/4! + (5 + x^2)*(4 + 2*x^2)*(3 + 3*x^2)*(2 + 4*x^2)*(1 + 5*x^2)*x^5/5! + ...
Also,
A(x)^2/(1 + x*A(x)) = 1 + (1 + 2*x^2)*x + (2 + 2*x^2)*(1 + 3*x^2)*x^2/2! + (3 + 2*x^2)*(2 + 3*x^2)*(1 + 4*x^2)*x^3/3! + (4 + 2*x^2)*(3 + 3*x^2)*(2 + 4*x^2)*(1 + 5*x^2)*x^4/4! + (5 + 2*x^2)*(4 + 3*x^2)*(3 + 4*x^2)*(2 + 5*x^2)*(1 + 6*x^2)*x^5/5! + ...
And,
A(x)^3/((1 + x*A(x))*(1 + x^3*A(x))) = 1 + (2 + 2*x^2)*x + (3 + 2*x^2)*(2 + 3*x^2)*x^2/2! + (4 + 2*x^2)*(3 + 3*x^2)*(2 + 4*x^2)*x^3/3! + (5 + 2*x^2)*(4 + 3*x^2)*(3 + 4*x^2)*(2 + 5*x^2)*x^4/4! + (6 + 2*x^2)*(5 + 3*x^2)*(4 + 4*x^2)*(3 + 5*x^2)*(2 + 6*x^2)*x^5/5! + ...
RELATED SERIES.
A(x)/(1 + x*A(x)) = 1 + 6*x^3/3! + 12*x^4/4! + 40*x^5/5! + 900*x^6/6! + 7728*x^7/7! + 68880*x^8/8! + 1031616*x^9/9! + ...
A(x)/(1 + x^3*A(x)) = 1 + x + 2*x^2/2! + 6*x^3/3! + 36*x^4/4! + 280*x^5/5! + 2460*x^6/6! + 25368*x^7/7! + 310128*x^8/8! + 4333824*x^9/9! + ...
where ( A(x)/(1 + x^3*A(x)) )^(x^2) = A(x)/(1 + x*A(x)).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..300

Crossrefs

Cf. A316370, A316700, A316701.

Programs

Mathematica

nmax = 20; CoefficientList[Series[Sum[(x^k*(x^2 - 1)^k * Pochhammer[(k + x^2)/(x^2 - 1), k])/k!, {k, 0, nmax}], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jul 15 2018 *)

PARI

{a(n) = my(A=1); A = sum(m=0,n, x^m/m! * prod(k=1,m, m+1-k + k*x^2 +x*O(x^n))); n!*polcoeff(A,n)}
for(n=0,30, print1(a(n),", "))

Formula

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:

(1) A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x^2.

(2) Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)*x^2 = A(x)^(p+q+1) / ( (1 + x*A(x))^q * (1 + x^3*A(x))^p ), for fixed p and q.

(3) A(x)/(1 + x*A(x)) = ( A(x)/(1 + x^3*A(x)) )^(x^2).

a(n) ~ 3^((n+1)/2) * n! / sqrt(Pi*log(3)*n). - Vaclav Kotesovec, Jul 15 2018