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a(n) = round(n^n/n!) where round(1/2)=0.
1

%I #28 Jan 12 2025 14:26:58

%S 1,1,2,4,11,26,65,163,416,1068,2756,7148,18614,48639,127463,334865,

%T 881658,2325751,6145597,16263866,43099804,114356611,303761260,

%U 807692035,2149632061,5726042115,15264691107,40722913454,108713644516,290404350963,776207020880

%N a(n) = round(n^n/n!) where round(1/2)=0.

%C We have two versions of this sequence, this and A235496, because there is no universal agreement on how to round a number like 9/2. - _N. J. A. Sloane_, Dec 13 2015

%C According to Stirling's asymptotic formula for n!, we have lim_{n->infinity} a(n+1)/a(n) = e.

%F For n>=1, round(A000312(n)/A000142(n)).

%t Table[Round[n^n/n!], {n, 30}] (* _Wesley Ivan Hurt_, Apr 08 2014 *)

%Y Cf. A000142, A000312.

%Y Essentially the same as A235496. - _Altug Alkan_, Dec 13 2015

%K nonn,easy,changed

%O 0,3

%A _Vladimir Shevelev_, Apr 08 2014

%E Name clarified by _Sean A. Irvine_, Jan 12 2025