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%I #11 Jun 23 2023 07:28:12
%S 1,2,52,3272,382672,71819552,19755648832,7489898916992,
%T 3743721038908672,2385494267756237312,1887436919680269939712,
%U 1815491288416066631616512,2086364959404184854563049472,2823211429546048668686123343872,4443155724532239407325655263035392
%N a(n) = (1/2) * Sum_{k>=0} (k*(k - 1))^n / 2^k.
%F a(n) = Sum_{k=0..n} (-1)^k * binomial(n,k) * A000670(2*n-k).
%F a(n) = 2 * A080163(n) for n > 0. - _Hugo Pfoertner_, Jan 23 2021
%F a(n) = A122101(2*n,n). - _Alois P. Heinz_, Jun 23 2023
%t Table[(1/2) Sum[(k (k - 1))^n/2^k, {k, 0, Infinity}], {n, 0, 14}]
%t Table[(1/2) Sum[(-1)^k Binomial[n, k] HurwitzLerchPhi[1/2, k - 2 n, 0], {k, 0, n}], {n, 0, 14}]
%Y Cf. A000670, A020556, A052841, A080163, A098696, A122101, A249938.
%K nonn
%O 0,2
%A _Ilya Gutkovskiy_, Jan 23 2021
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