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A330188 a(n) = round(n^arctan(n!)). 0

%I #47 Jan 17 2024 06:58:05

%S 1,2,5,8,12,17,21,26,32,37,43,50,56,63,70,78,86,94,102,111,119,128,

%T 138,147,157,167,177,188,198,209,220,231,243,254,266,278,291,303,316,

%U 328,341,355,368,382,395,409,423,437,452,466,481,496,511,526,542,557,573

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

%C Due to the limited range of the inverse tangent function, n^arctan(n!) approaches n^(Pi/2), but never reaches it.

%C It appears that a(n) = round(n^(Pi/2)) for all n > 5. - _Jon E. Schoenfield_, Dec 07 2019

%F a(n) = round(n^arctan(n!)).

%e a(1) = 1 because 1^arctan(1!) = 1^arctan(1) = 1^0.785398163... --> 1;

%e a(2) = 2 because 2^arctan(2!) = 2^arctan(2) = 2^1.1071487... = 2.1541948... --> 2;

%e a(3) = 5 because 3^arctan(3!) = 3^arctan(6) = 3^1.4056476... = 4.6845121... --> 5.

%t a[n_] := Round[n^ArcTan[n!]]; Array[a, 57] (* _Amiram Eldar_, Dec 06 2019 *)

%o (JavaScript)

%o var list = [];

%o function factorial(b) {

%o var h = 1;

%o for (var i = 1; i <= b; i++) {

%o h=h*i;

%o }

%o return(h);

%o }

%o for (var i = 1; i < 50; i++) {

%o var g = Math.pow(i,Math.atan(factorial(i)));

%o appendItem(list,Math.round(g));

%o }

%o console.log(list);

%o (PARI) a(n) = round(n^atan(n!)); \\ _Michel Marcus_, Jan 17 2024

%K nonn

%O 1,2

%A _Sebastian F. Orellana_, Dec 04 2019

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Last modified July 24 18:12 EDT 2024. Contains 374585 sequences. (Running on oeis4.)