OFFSET
1,2
COMMENTS
Due to the limited range of the inverse tangent function, n^arctan(n!) approaches n^(Pi/2), but never reaches it.
It appears that a(n) = round(n^(Pi/2)) for all n > 5. - Jon E. Schoenfield, Dec 07 2019
FORMULA
a(n) = round(n^arctan(n!)).
EXAMPLE
a(1) = 1 because 1^arctan(1!) = 1^arctan(1) = 1^0.785398163... --> 1;
a(2) = 2 because 2^arctan(2!) = 2^arctan(2) = 2^1.1071487... = 2.1541948... --> 2;
a(3) = 5 because 3^arctan(3!) = 3^arctan(6) = 3^1.4056476... = 4.6845121... --> 5.
MATHEMATICA
a[n_] := Round[n^ArcTan[n!]]; Array[a, 57] (* Amiram Eldar, Dec 06 2019 *)
PROG
(JavaScript)
var list = [];
function factorial(b) {
var h = 1;
for (var i = 1; i <= b; i++) {
h=h*i;
}
return(h);
}
for (var i = 1; i < 50; i++) {
var g = Math.pow(i, Math.atan(factorial(i)));
appendItem(list, Math.round(g));
}
console.log(list);
(PARI) a(n) = round(n^atan(n!)); \\ Michel Marcus, Jan 17 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Sebastian F. Orellana, Dec 04 2019
STATUS
approved