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a(n) = round(n^n/n!).
2

%I #33 Sep 08 2022 08:46:06

%S 1,1,2,5,11,26,65,163,416,1068,2756,7148,18614,48639,127463,334865,

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

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

%N a(n) = round(n^n/n!).

%C We have two versions of this sequence, this and A240571, because there is no universal agreement on how to round a number like 9/2. - _N. J. A. Sloane_, Dec 13 2015

%H Vincenzo Librandi, <a href="/A235496/b235496.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = Round(A000312(n)/A000142(n)).

%p a:= n-> round(n^n/n!):

%p seq(a(n), n=0..32); # _Alois P. Heinz_, Dec 13 2015

%t Table[Floor[n^n/n! + 1/2], {n, 40}]

%o (Magma) [Round(n^n/Factorial(n)): n in [1..40]];

%o (PARI) s=[]; for(n=1, 30, s=concat(s, round(n^n/n!))); s \\ _Colin Barker_, Jan 19 2014

%Y Cf. A000142, A000312, A055775, A073225.

%Y See A240571 for another version.

%K nonn,easy

%O 0,3

%A _Vincenzo Librandi_, Jan 15 2014