A259748 a(n) = (Sum_{0<x<y<n} x*y) mod n.

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, 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, 0, 0, 25, 0, 31, 42, 16, 0, 44, 0, 51, 46, 35, 0, 66, 0, 0, 50

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

Cf. A000914,

A259749 (n such that a(n)=0),

A259750 (n such that n/a(n)=2),

A259751 (n such that n/a(n)=4),

A259752 (n such that n/a(n)=6),

A073762 (n such that n/a(n)=12/5),

A259754 (n such that n/a(n)=3/2),

A259755 (n such that n/a(n)=4/3),

A008606 (n such that n/a(n)=12/11).

Sequence in context: A215151 A305806 A064722 * A357728 A357681 A357720

Adjacent sequences: A259745 A259746 A259747 * A259749 A259750 A259751

Keyword

nonn,easy

Author

José María Grau Ribas, Jul 04 2015

Status

approved