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a(n) = 3^(n-1) - ceiling(n^n/n!).
1

%I #11 Sep 08 2022 08:45:29

%S 0,1,4,16,54,178,565,1770,5493,16927,51901,158533,482802,1466859,

%T 4448104,13467249,40720970,122994566,371156622,1119161662,3372427789,

%U 10156591942,30573367574,91993546765,276703494365,832023918335,2501142914874,7516883840470

%N a(n) = 3^(n-1) - ceiling(n^n/n!).

%C Theorem: 3^(n-1) > n^n/n! for n >= 3.

%D D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 193, 3.1.21.

%H Robert Israel, <a href="/A127634/b127634.txt">Table of n, a(n) for n = 1..2094</a>

%p seq(3^(n-1)-ceil(n^n/n!),n=1..50); # _Robert Israel_, Jul 06 2017

%t Table[3^(n-1) - Ceiling[n^n / n!], {n, 30}] (* _Vincenzo Librandi_, Jul 06 2017 *)

%o (PARI) a(n) = 3^(n-1) - ceil(n^n/n!); \\ _Michel Marcus_, Jul 06 2017

%o (Magma) [3^(n-1)-Ceiling(n^n/Factorial(n)): n in [1..30]]; // _Vincenzo Librandi_, Jul 06 2017

%Y Cf. A000244, A073225.

%K nonn

%O 1,3

%A _N. J. A. Sloane_, Apr 03 2007