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%I #5 Jun 29 2019 08:58:44
%S 1,3,75,545835,5315654681981355,
%T 23545154085734896649184490637144855476395,
%U 1408419189834457368564204829523128565178390518208834734403870775971211916384631776000167564122146475
%N a(n) = Sum_{k>=0} k^(2^n)/2^(k+1).
%F a(n) = Sum_{k=0..2^n} k!*Stirling2(2^n,k).
%F a(n) = A000670(A000079(n)).
%t Table[Sum[k^(2^n)/2^(k + 1), {k, 0, Infinity}], {n, 0, 6}]
%t Table[Sum[k! StirlingS2[2^n, k], {k, 0, 2^n}], {n, 0, 6}]
%Y Cf. A000079, A000670, A008277, A135085.
%K nonn
%O 0,2
%A _Ilya Gutkovskiy_, Jun 29 2019
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