A167139 G.f.: Sum_{n>=0} A005649(n)^2 * log(1+x)^n/n! where 1/(1-x)^2 = Sum_{n>=0} A005649(n)*log(1+x)^n/n!.

1, 4, 30, 292, 3497, 49488, 806504, 14860032, 305261640, 6914828176, 171186477632, 4597513706496, 133116705145408, 4133143450593536, 136981118139314688, 4826352390162440704, 180139085757269111824

Offset: 0

Author

Paul D. Hanna, Nov 03 2009

Keywords

• nonn

Comments

Conjecture: For all integers m > 0, Sum_{n>=0} L(n)^m * log(1+x)^n/n! is an integer series whenever Sum_{n>=0} L(n)*log(1+x)^n/n! is an integer series.

Examples

			G.f.: A(x) = 1 + 4*x + 30*x^2 + 292*x^3 + 3497*x^4 + 49488*x^5 + ...
Illustrate A(x) = Sum_{n>=0} A005649(n)^2 * log(1+x)^n/n!:
A(x) = 1 + 2^2*log(1+x) + 8^2*log(1+x)^2/2! + 44^2*log(1+x)^3/3! + 308^2*log(1+x)^4/4! + 2612^2*log(1+x)^5/5! + ... + A005649(n)^2*log(1+x)^n/n! + ...
where the g.f. of A005649 is 1/(2 - exp(x))^2:
1/(1-x)^2 = 1 + 2*log(1+x) + 8*log(1+x)^2/2! + 44*log(1+x)^3/3! + 308*log(1+x)^4/4! + 2612*log(1+x)^5/5! + ... + A005649(n)*log(1+x)^n/n! + ...
		

Crossrefs

Cf. A167138, A005649.

Programs

PARI

{A005649(n)=sum(k=0,n,(k+1)*stirling(n, k, 2)*k!)}
{a(n)=polcoef(sum(m=0,n,A005649(m)^2*log(1+x+x*O(x^n))^m/m!),n)}

Formula

a(n) = (1/n!)*Sum_{k=0..n} Stirling1(n,k)*A005649(k)^2, cf. A101370. - Vladeta Jovovic, Nov 09 2009