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%I #12 Mar 13 2023 07:20:40
%S 1,1,5,55,961,24101,818821,36053515,1984670465,132825475081,
%T 10583425959301,988018789759871,106673677280748865,
%U 13172700275176482925,1842428769970603518341,289406832942160060794451,50677793314733587473331201,9829328870566195730521433105
%N a(n) = n! * [x^n] exp(x/(1 - x)^n).
%C Conjecture: a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. - _Peter Bala_, Mar 12 2023
%H Seiichi Manyama, <a href="/A293013/b293013.txt">Table of n, a(n) for n = 0..274</a>
%F a(n) = A293012(n,n).
%t Table[n! SeriesCoefficient[Exp[x/(1 - x)^n] , {x, 0, n}], {n, 0, 17}]
%Y Main diagonal of A293012. Cf. A361281.
%K nonn,easy
%O 0,3
%A _Ilya Gutkovskiy_, Sep 28 2017