OFFSET
1,3
COMMENTS
{a(n)/n: n=1,2,...} = {0, 1/6, 1/4, 5/12, 1/2, 2/3, 3/4, 11/12}.
From Danny Rorabaugh, Oct 22 2015: (Start)
a(n)/n = 0 iff n mod 24 = 1,2,5,7,10,11,13,17,19,23 (A259749);
a(n)/n = 1/6 iff n mod 24 = 6 (A259752);
a(n)/n = 1/4 iff n mod 24 = 8,16 (A259751);
a(n)/n = 5/12 iff n mod 24 = 12 (A073762);
a(n)/n = 1/2 iff n mod 24 = 14,22 (A259750);
a(n)/n = 2/3 iff n mod 24 = 3,9,15,18,21 (A259754);
a(n)/n = 3/4 iff n mod 24 = 4,20 (A259755);
a(n)/n = 11/12 iff n mod 24 = 0 (A008606).
(End)
LINKS
Danny Rorabaugh, Table of n, a(n) for n = 1..24000
Danny Rorabaugh, Proof of a(n)/n values for A259748
FORMULA
a(n) = A000914(n) mod n = (1/24)*(-1 + n)*n*(1 + n)*(2 + 3*n) mod n.
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
MATHEMATICA
A[n_]:=Sum[a b, {a, 1, n}, {b, a+1, n}]; Table[Mod[A[n], n], {n, 1, 122}]
PROG
(PARI) vector(100, n, ((n-1)*n*(n+1)*(3*n+2)/24) % n) \\ Altug Alkan, Oct 22 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
José María Grau Ribas, Jul 04 2015
STATUS
approved