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%I #5 Sep 28 2016 23:59:55
%S 1,1,7,17,56,199,890,4649,27817,195946,1684398,17397323,208799982,
%T 2932164012,49785808832,1022745137705,24671296028079,695270673553051,
%U 23526126768837873,965093874912658722,46827415587504280547,2655503102769481320544,179856174616910379655073,14761130793635395568878091,1439881917495260610082685956,164363140573098989525137162900,22322323085863965044351721067969
%N L.g.f.: Sum_{n>=1} [ Sum_{k>=1} k^n * x^(2*k-1) ]^n / n.
%C L.g.f. equals the logarithm of the g.f. of A276906.
%H Paul D. Hanna, <a href="/A276907/b276907.txt">Table of n, a(n) for n = 1..200</a>
%F L.g.f.: Sum_{n>=1} [ Sum_{k=1..n} A008292(n,k) * x^(2*k-1) / (1-x^2)^(n+1) ]^n / n, where A008292 are the Eulerian numbers.
%e L.g.f.: A(x) = x + x^2/2 + 7*x^3/3 + 17*x^4/4 + 56*x^5/5 + 199*x^6/6 + 890*x^7/7 + 4649*x^8/8 + 27817*x^9/9 + 195946*x^10/10 + 1684398*x^11/11 + 17397323*x^12/12 +...
%e such that A(x) equals the series:
%e A(x) = Sum_{n>=1} (x + 2^n*x^3 + 3^n*x^5 +...+ k^n*x^(2*k-1) +...)^n/n.
%e This logarithmic series can be written using the Eulerian numbers like so:
%e A(x) = x/(1-x^2)^2 + (x + x^3)^2/(1-x^2)^6/2 + (x + 4*x^3 + x^5)^3/(1-x^2)^12/3 + (x + 11*x^3 + 11*x^5 + x^7)^4/(1-x^2)^20/4 + (x + 26*x^3 + 66*x^5 + 26*x^7 + x^9)^5/(1-x^2)^30/5 + (x + 57*x^3 + 302*x^5 + 302*x^7 + 57*x^9 + x^11)^6/(1-x^2)^42/6 +...+ [ Sum_{k=1..n} A008292(n,k) * x^(2*k-1) ]^n / (1-x^2)^(n*(n+1))/n +...
%e where
%e exp(A(x)) = 1 + x + x^2 + 3*x^3 + 7*x^4 + 18*x^5 + 53*x^6 + 188*x^7 + 799*x^8 + 4001*x^9 + 24050*x^10 + 179248*x^11 + 1639637*x^12 +...+ A276906(n)*x^n +...
%o (PARI) {a(n) = n * polcoeff( sum(m=1, n, sum(k=1, n, k^m * x^(2*k-1) +x*O(x^n))^m/m ), n)}
%o for(n=1, 30, 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, m, A008292(m, k) * x^(2*k-1)/(1-x^2 +Oxn)^(m+1) )^m / m ); n*polcoeff(A, n)}
%o for(n=1, 30, print1(a(n), ", "))
%Y Cf. A276906, A276750.
%K nonn
%O 1,3
%A _Paul D. Hanna_, Sep 28 2016
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