|
%I #26 Aug 30 2021 21:40:28
%S 0,1,12,45,225,396,960,1377,2541,5292,6525,11340,15600,18081,23805,
%T 34476,47937,53100,70785,84525,92016,117117,136161,168432,218880,
%U 247500,262701,294945,312012,348096,496125,545025,624240,652257,804972,838125,943020
%N a(n) = ((p-1)^3 - (p-1)^2)/4 where p is the n-th prime.
%H Alois P. Heinz, <a href="/A331764/b331764.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeSums.html">Prime Sum</a>
%F 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_
%F a(n) == 0 (mod 3) for n >= 3. - _Hugo Pfoertner_, Aug 23 2021
%p a:= n-> (p-> ((p-1)^3-(p-1)^2)/4)(ithprime(n)):
%p seq(a(n), n=1..40); # _Alois P. Heinz_, Feb 05 2020
%t Table[((Prime[n] - 1)^3 - (Prime[n] - 1)^2)/4, {n, 20}] (* _Eric W. Weisstein_, Aug 22 2021 *)
%t Table[((Prime[n] - 2) (Prime[n] - 1)^2)/4, {n, 20}] (* _Eric W. Weisstein_, Aug 22 2021 *)
%t Table[Times @@ (Prime[n] - {1, 1, 2})/4, {n, 20}] (* _Eric W. Weisstein_, Aug 22 2021 *)
%t Table[Sum[Floor[i j/Prime[n]], {i, Prime[n] - 1}, {j, Prime[n] - 1}], {n, 20}] (* _Eric W. Weisstein_, Aug 22 2021 *)
%K nonn
%O 1,3
%A _N. J. A. Sloane_, Feb 05 2020 following a suggestion from _Jean-Claude Babois_.
|
