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%I #29 Nov 27 2022 02:41:42
%S 1,6,150,9366,1091670,204495126,56183135190,21282685940886,
%T 10631309363962710,6771069326513690646,5355375592488768406230,
%U 5149688839606380769088406,5916558242148290945301297750,8004451519688336984972255078166,12595124129900132067036747870669270
%N a(n) = Sum_{k>=1} k^(2*n)/(2^k).
%H Vincenzo Librandi, <a href="/A227044/b227044.txt">Table of n, a(n) for n = 0..100</a>
%F a(n) ~ (2n)!/(log(2))^(2*n+1).
%F a(n) = Sum_{k=0..2*n} (-2)^k * k! * Stirling2(2*n, k). - _Paul D. Hanna_, Apr 15 2018
%F a(n) = A000629(2*n). - _Christian Krause_, Nov 22 2022
%t Table[Sum[k^(2*n)/(2^k), {k, 1, Infinity}], {n, 0, 20}]
%t a[n_] := PolyLog[-2 n, 1/2]; a[0] = 1; Array[a, 15, 0] (* _Peter Luschny_, Sep 06 2020 *)
%o (PARI) {a(n) = sum(k=0, 2*n, (-2)^k * k! * stirling(2*n, k,2) )}
%o for(n=0, 20, print1(a(n), ", "))
%Y Bisection of A000629.
%Y Cf. A080163.
%K nonn
%O 0,2
%A _Vaclav Kotesovec_, Jun 29 2013
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