A331764 a(n) = ((p-1)^3 - (p-1)^2)/4 where p is the n-th prime.

0, 1, 12, 45, 225, 396, 960, 1377, 2541, 5292, 6525, 11340, 15600, 18081, 23805, 34476, 47937, 53100, 70785, 84525, 92016, 117117, 136161, 168432, 218880, 247500, 262701, 294945, 312012, 348096, 496125, 545025, 624240, 652257, 804972, 838125, 943020

Offset: 1

Author

N. J. A. Sloane, Feb 05 2020 following a suggestion from Jean-Claude Babois

Keywords

• nonn

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Prime Sum

Programs

Maple

a:= n-> (p-> ((p-1)^3-(p-1)^2)/4)(ithprime(n)):
seq(a(n), n=1..40);  # Alois P. Heinz, Feb 05 2020

Mathematica

Table[((Prime[n] - 1)^3 - (Prime[n] - 1)^2)/4, {n, 20}] (* Eric W. Weisstein, Aug 22 2021 *)
Table[((Prime[n] - 2) (Prime[n] - 1)^2)/4, {n, 20}] (* Eric W. Weisstein, Aug 22 2021 *)
Table[Times @@ (Prime[n] - {1, 1, 2})/4, {n, 20}] (* Eric W. Weisstein, Aug 22 2021 *)
Table[Sum[Floor[i j/Prime[n]], {i, Prime[n] - 1}, {j, Prime[n] - 1}], {n, 20}] (* Eric W. Weisstein, Aug 22 2021 *)

Formula

Theorem: a(n) = Sum_{i=1..p-1, j=1..p-1} floor(i*j/p). The proof is based on the formula for p-g-c-d of Marcelo Polezzi. - Jean-Claude Babois

a(n) == 0 (mod 3) for n >= 3. - Hugo Pfoertner, Aug 23 2021