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a(n) = Sum_{k=1..n} k! * k^(2*n-1) * Stirling1(n,k).
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%I #12 Jun 19 2024 09:28:09

%S 0,1,15,1268,317294,175542694,181641609214,315309390376056,

%T 850661260866748728,3370191684116333977872,18768704088141613880906736,

%U 141902519646656406912522712848,1415862822521619228707500717132224,18210234893009450819658863637633454608

%N a(n) = Sum_{k=1..n} k! * k^(2*n-1) * Stirling1(n,k).

%F E.g.f.: Sum_{k>=1} log(1 + k^2*x)^k / k.

%t nmax=13; Range[0,nmax]!CoefficientList[Series[Sum[(Log[1 + k^2*x])^k / k,{k,nmax}],{x,0,nmax}],x] (* _Stefano Spezia_, Jun 19 2024 *)

%o (PARI) a(n) = sum(k=1, n, k!*k^(2*n-1)*stirling(n, k, 1));

%Y Cf. A220179, A242228, A373856.

%Y Cf. A351133, A351183.

%K nonn

%O 0,3

%A _Seiichi Manyama_, Jun 19 2024