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%I #9 Sep 18 2016 15:49:29
%S 1,9,76,1157,33291,1792296,196919213,39766253741,16931726147956,
%T 13298466280839329,22076711237844558263,69166686377284889199104,
%U 448760359479425463648647769,5685081590883001302122022078913,144528951819771627855280850227089996,7431791795502279858136165452572662669213,743200333842768450767851829731370148558347843,154769006272445896954868694741314742556915451805336
%N L.g.f.: Sum_{n>=1} [ Sum_{k>=1} k^(2*n) * x^k ]^n / n.
%C L.g.f. equals the logarithm of the g.f. of A276752.
%H Paul D. Hanna, <a href="/A276754/b276754.txt">Table of n, a(n) for n = 1..100</a>
%F L.g.f.: Sum_{n>=1} [ Sum_{k=1..2*n-1} A008292(2*n,k) * x^k / (1-x)^(2*n+1) ]^n / n, where A008292 are the Eulerian numbers.
%e L.g.f.: A(x) = x + 9*x^2/2 + 76*x^3/3 + 1157*x^4/4 + 33291*x^5/5 + 1792296*x^6/6 + 196919213*x^7/7 + 39766253741*x^8/8 + 16931726147956*x^9/9 + 13298466280839329*x^10/10 +...
%e such that A(x) equals the series:
%e A(x) = Sum_{n>=1} (x + 2^(2*n)*x^2 + 3^(2*n)*x^3 +...+ k^(2*n)*x^k +...)^n/n.
%e This logarithmic series can be written using the Eulerian numbers like so:
%e A(x) = (x + x^2)/(1-x)^3 + (x + 11*x^2 + 11*x^3 + x^4)^2/(1-x)^10/2 + (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)^3/(1-x)^21/3 + (x + 247*x^2 + 4293*x^3 + 15619*x^4 + 15619*x^5 + 4293*x^6 + 247*x^7 + x^8)^4/(1-x)^36/4 + (x + 1013*x^2 + 47840*x^3 + 455192*x^4 + 1310354*x^5 + 1310354*x^6 + 455192*x^7 + 47840*x^8 + 1013*x^9 + x^10)^5/(1-x)^55/5 + (x + 4083*x^2 + 478271*x^3 + 10187685*x^4 + 66318474*x^5 + 162512286*x^6 + 162512286*x^7 + 66318474*x^8 + 10187685*x^9 + 478271*x^10 + 4083*x^11 + x^12)^6/(1-x)^78/6 +...+ [ Sum_{k=1..2*n} A008292(2*n,k) * x^k ]^n / (1-x)^(2*n^2+n) /n +...
%e where
%e exp(A(x)) = 1 + x + 5*x^2 + 30*x^3 + 327*x^4 + 7085*x^5 + 307280*x^6 + 28472653*x^7 + 5000661017*x^8 + 1886425568702*x^9 + 1331753751874235*x^10 +...+ A276752(n)*x^n +...
%o (PARI) {a(n) = n * polcoeff( sum(m=1, n, sum(k=1, n, k^(2*m)*x^k +x*O(x^n))^m/m ), n)}
%o for(n=1, 20, print1(a(n), ", "))
%o (PARI) {A008292(n, k) = sum(j=0, k, (-1)^j * (k-j)^n * binomial(n+1, j))}
%o {a(n) = my(A=1, Oxn=x*O(x^n)); A = sum(m=1, n+1, sum(k=1, 2*m, A008292(2*m, k)*x^k/(1-x +Oxn)^(2*m+1) )^m / m ); n * polcoeff(A, n)}
%o for(n=1, 20, print1(a(n), ", "))
%Y Cf. A156170, A276753, A008292.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Sep 17 2016