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%I #54 Oct 23 2015 03:53:07
%S 0,0,2,3,0,1,0,2,6,0,0,5,0,7,10,4,0,12,0,15,14,11,0,22,0,0,18,21,0,5,
%T 0,8,22,0,0,15,0,19,26,10,0,28,0,33,30,23,0,44,0,0,34,39,0,9,0,14,38,
%U 0,0,25,0,31,42,16,0,44,0,51,46,35,0,66,0,0,50
%N a(n) = (Sum_{0<x<y<n} x*y) mod n.
%C {a(n)/n: n=1,2,...} = {0, 1/6, 1/4, 5/12, 1/2, 2/3, 3/4, 11/12}.
%C From _Danny Rorabaugh_, Oct 22 2015: (Start)
%C a(n)/n = 0 iff n mod 24 = 1,2,5,7,10,11,13,17,19,23 (A259749);
%C a(n)/n = 1/6 iff n mod 24 = 6 (A259752);
%C a(n)/n = 1/4 iff n mod 24 = 8,16 (A259751);
%C a(n)/n = 5/12 iff n mod 24 = 12 (A073762);
%C a(n)/n = 1/2 iff n mod 24 = 14,22 (A259750);
%C a(n)/n = 2/3 iff n mod 24 = 3,9,15,18,21 (A259754);
%C a(n)/n = 3/4 iff n mod 24 = 4,20 (A259755);
%C a(n)/n = 11/12 iff n mod 24 = 0 (A008606).
%C (End)
%H Danny Rorabaugh, <a href="/A259748/b259748.txt">Table of n, a(n) for n = 1..24000</a>
%H Danny Rorabaugh, <a href="/A259748/a259748_1.pdf">Proof of a(n)/n values for A259748</a>
%F a(n) = A000914(n) mod n = (1/24)*(-1 + n)*n*(1 + n)*(2 + 3*n) mod n.
%F a(24k) = 22k; a(24k+1) = 0; a(24k+2) = 0; a(24k+3) = 16k+2; a(24k+4) = 18k+3; a(24k+5) = 0; a(24k+6) = 4k+1, a(24k+7) = 0; a(24k+8) = 6k+2; a(24k+9) = 16k+6; a(24k+10) = 0; a(24k+11) = 0; a(24k+12) = 10k+5; a(24k+13) = 0; a(24k+14) = 12k+7; a(24k+15) = 16k+10; a(24k+16) = 6k+4; a(24k+17) = 0; a(24k+18) = 16k+12; a(24k+19) = 0; a(24k+20) = 18k+15; a(24k+21) = 16k+14; a(24k+22) = 12k+11; a(24k+23) = 0. - _Danny Rorabaugh_, Oct 22 2015
%t A[n_]:=Sum[a b,{a,1,n},{b,a+1,n}];Table[Mod[A[n],n],{n,1,122}]
%o (PARI) vector(100, n, ((n-1)*n*(n+1)*(3*n+2)/24) % n) \\ _Altug Alkan_, Oct 22 2015
%Y Cf. A000914,
%Y A259749 (n such that a(n)=0),
%Y A259750 (n such that n/a(n)=2),
%Y A259751 (n such that n/a(n)=4),
%Y A259752 (n such that n/a(n)=6),
%Y A073762 (n such that n/a(n)=12/5),
%Y A259754 (n such that n/a(n)=3/2),
%Y A259755 (n such that n/a(n)=4/3),
%Y A008606 (n such that n/a(n)=12/11).
%K nonn,easy
%O 1,3
%A _José María Grau Ribas_, Jul 04 2015
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