A259748 a(n) = (Sum_{0

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

Author

José María Grau Ribas, Jul 04 2015

Keywords

• nonn
• easy

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

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).

Programs

Mathematica

A[n_]:=Sum[a b,{a,1,n},{b,a+1,n}];Table[Mod[A[n],n],{n,1,122}]

PARI

vector(100, n, ((n-1)*n*(n+1)*(3*n+2)/24) % n) \\ Altug Alkan, Oct 22 2015

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