%I #9 Jun 03 2022 06:06:02
%S 0,1,7,56,538,6124,81048,1226112,20902992,396857376,8308373760,
%T 190212376320,4728556327680,126865966625280,3654264347274240,
%U 112484501485977600,3685202487258163200,128039255560187596800
%N Weighted sum of the harmonic numbers.
%F a(n) = 2*n*a(n-1) + (n-1)!*(2^n-1); a(0)=0, a(1)=1. a(n)=n! * sum(j=1, n, binomial(n, j)*H(j)), where H(j)=sum(k=1, j, 1/k).
%F E.g.f.: log((2*x-1)/(x-1))/(2*x-1). a(n) = n!*Sum_{k=1..n} (-1)^(k+1)*2^(n-k)*binomial(n, k)/k. a(n) = n!*Sum_{k=1..n} 2^(n-k)*(2^k-1)/k. - _Vladeta Jovovic_, Aug 12 2005
%F a(n) ~ n! * log(n) * 2^n * (1 + (gamma-log(2))/log(n)), where gamma is the Euler-Mascheroni constant A001620. - _Vaclav Kotesovec_, Jun 03 2022
%o (PARI) H(n)=sum(j=1,n,1/j); a(n)=n!*sum(j=1,n,binomial(n,j)*H(j))
%Y Cf. A103213, A068102.
%K easy,nonn
%O 0,3
%A Nicholas C. Singer (nsinger2(AT)cox.net), Oct 02 2003