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a(n) = Sum_{k>=0} k^(2*n+1)/2^(k+1).
0

%I #7 Sep 25 2019 04:51:16

%S 1,13,541,47293,7087261,1622632573,526858348381,230283190977853,

%T 130370767029135901,92801587319328411133,81124824998504073881821,

%U 85438451336745709294580413,106697365438475775825583498141,155897763918621623249276226253693,263478385263023690020893329044576861

%N a(n) = Sum_{k>=0} k^(2*n+1)/2^(k+1).

%F a(n) = Sum_{k=0..2*n+1} k!*Stirling2(2*n+1,k).

%F a(n) = A000670(2*n+1).

%F a(n) ~ sqrt(Pi) * 2^(2*n + 1) * n^(2*n + 3/2) / (exp(2*n) * (log(2))^(2*n + 2)). - _Vaclav Kotesovec_, Sep 25 2019

%t Table[Sum[k^(2 n + 1)/2^(k + 1), {k, 0, Infinity}], {n, 0, 14}]

%t Table[Sum[k! StirlingS2[2 n + 1, k], {k, 0, 2 n + 1}], {n, 0, 14}]

%Y Cf. A000670, A008277, A099977, A249938.

%K nonn

%O 0,2

%A _Ilya Gutkovskiy_, Jun 29 2019