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G.f.: Sum_{n>=0} A155585(n)^2 * log(1/(1-2*x))^n/n!, where 1/(1-2*x+2*x^2) = Sum_{n>=0} A155585(n)*log(1/(1-2*x))^n/n!.
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%I #2 Mar 30 2012 18:37:20

%S 1,2,2,8,20,112,432,3200,16704,154688,1017920,11333888,90011264,

%T 1172330496,10908526592,162802835456,1737036006400,29235365490688,

%U 351847501606912,6593866787569664,88364197074231296,1825016315965767680

%N G.f.: Sum_{n>=0} A155585(n)^2 * log(1/(1-2*x))^n/n!, where 1/(1-2*x+2*x^2) = Sum_{n>=0} A155585(n)*log(1/(1-2*x))^n/n!.

%C Note that A155585(n) = 2^n E_{n}(1) where E_{n}(x) are the Euler polynomials; e.g.f. of A155585 is exp(x)/cosh(x).

%C CONJECTURE: For all integer 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.

%e G.f.: A(x) = 1 + 2*x + 2*x^2 + 8*x^3 + 20*x^4 + 112*x^5 + 432*x^6 +...

%e Illustrate A(x) = Sum_{n>=0} A155585(n)^2*log(1/(1-2*x))^n/n!:

%e A(x) = 1 - log(1-2*x) - 2^2*log(1-2*x)^3/3! - 16^2*log(1-2*x)^5/5! - 272^2*log(1-2*x)^7/7! - 7936^2*log(1-2*x)^9/9! +...+ A155585(n)^2*[ -log(1-2x)]^n/n! +...

%e where:

%e 1/((1-x)^2 + x^2) = 1 - log(1-2*x) + 2*log(1-2*x)^3/3! - 16*log(1-2*x)^5/5! + 272*log(1-2*x)^7/7! - 7936*log(1-2*x)^9/9! +...+ A155585(n)*[ -log(1-2x)]^n/n! +...

%o (PARI) {A155585(n)=if(n==0,1,bernfrac(n+1)*(2^(n+1)-1)*2^(n+1)/(n+1))}

%o {a(n)=polcoeff(sum(k=0,n,A155585(k)^2*log(1/(1-2*x +x*O(x^n)))^k/k!),n)}

%Y Cf. A155585, variants: A167141, A167139, A167138, A101370.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Nov 05 2009