A316704 E.g.f.: Sum_{n>=0} (2*x)^n/n! * Product_{k=1..n} (n+1-k) + k*x.

1, 2, 12, 108, 1312, 20320, 381408, 8420160, 213813248, 6139270656, 196691281920, 6956268042240, 269187901974528, 11313963679948800, 513251989767487488, 24995547184468008960, 1300702468667721646080, 72026879559935471124480, 4229000873160355032072192, 262425376836886982897958912, 17161024720479004010930503680, 1179556709319250468710226329600

Offset

0,2

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.

Links

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

Formula

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

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

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

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

a(n)/n! ~ c * d^n / sqrt(n), where d = 3.346513389529679772056152566067040813392... and c = 1.06774499146514892068040233... - Vaclav Kotesovec, Jul 18 2018

The constant d given above is the root of the equation d^(d-2) = 2^(d-1). - Vaclav Kotesovec, Jan 17 2024

Example

E.g.f.: A(x) = 1 + 2*x + 12*x^2/2! + 108*x^3/3! + 1312*x^4/4! + 20320*x^5/5! + 381408*x^6/6! + 8420160*x^7/7! + 213813248*x^8/8! + 6139270656*x^9/9! + 196691281920*x^10/10! + ...
such that
A(x) = 1 + (1+x)*(2*x) + (2 + x)*(1 + 2*x)*(2*x)^2/2! + (3 + x)*(2 + 2*x)*(1 + 3*x)*(2*x)^3/3! + (4 + x)*(3 + 2*x)*(2 + 3*x)*(1 + 4*x)*(2*x)^4/4! + (5 + x)*(4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*(2*x)^5/5! + ...
Also,
A(x)^2/(1 + 2*x*A(x)) = 1 + (1 + 2*x)*(2*x) + (2 + 2*x)*(1 + 3*x)*(2*x)^2/2! + (3 + 2*x)*(2 + 3*x)*(1 + 4*x)*(2*x)^3/3! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*(2*x)^4/4! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(1 + 6*x)*(2*x)^5/5! + ...
And,
A(x)^3/((1 + 2*x*A(x))*(1 + 2*x^2*A(x))) = 1 + (2 + 2*x)*(2*x) + (3 + 2*x)*(2 + 3*x)*(2*x)^2/2! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(2*x)^3/3! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(2*x)^4/4! + (6 + 2*x)*(5 + 3*x)*(4 + 4*x)*(3 + 5*x)*(2 + 6*x)*(2*x)^5/5! + ...
RELATED SERIES.
B(x) = sqrt( (1 + 2*x*A(x)) * (1 + 2*x^2*A(x)) ) = 1 + x + 5*x^2/2! + 45*x^3/3! + 513*x^4/4! + 7745*x^5/5! + 142485*x^6/6! + 3095421*x^7/7! + 77642145*x^8/8! + 2207145825*x^9/9! + ... + A316705(n)*x^n/n! + ...
where
B(x) = 1 + (1 + x)*x + (3 + x)*(1 + 3*x)*x^2/2! + (5 + x)*(3 + 3*x)*(1 + 5*x)*x^3/3! + (7 + x)*(5 + 3*x)*(3 + 5*x)*(1 + 7*x)*x^4/4! + (9 + x)*(7 + 3*x)*(5 + 5*x)*(3 + 7*x)*(1 + 9*x)*x^5/5! + ...

Prog

(PARI) /* Using the biexponential series */
{a(n) = my(A); A = sum(m=0, n, (2*x)^m/m! * prod(k=1, m, m+1-k + k*x +x*O(x^n))); n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* Using Functional Equation: */
{a(n) = my(A=1); for(i=1, n, A = ( (1 + 2*x*A)/(1 + 2*x^2*A +x*O(x^n))^x )^(1/(1-x)) ); n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A316370, A316705.

Sequence in context: A055897 A210997 A052563 * A228173 A218652 A194786

Adjacent sequences: A316701 A316702 A316703 * A316705 A316706 A316707

Keyword

nonn

Author

Paul D. Hanna, Jul 16 2018

Status

approved