A259751 Numbers that are congruent to {8, 16} mod 24.

8, 16, 32, 40, 56, 64, 80, 88, 104, 112, 128, 136, 152, 160, 176, 184, 200, 208, 224, 232, 248, 256, 272, 280, 296, 304, 320, 328, 344, 352, 368, 376, 392, 400, 416, 424, 440, 448, 464, 472, 488, 496, 512, 520, 536, 544, 560, 568, 584, 592, 608, 616, 632

Offset: 1

Author

José María Grau Ribas, Jul 06 2015

Keywords

• nonn
• easy

Comments

Original name: Numbers n such that n/A259748(n) = 4.

Links

• Danny Rorabaugh, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A000914, A001651.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259752, A259754, A259755.

Programs

Mathematica

A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/4 & ]

PARI

Vec(8*x*(1+x+x^2)/((1-x)^2*(1+x)) + O(x^100)) \\ Colin Barker, Aug 26 2016

Formula

A259748(a(n))/a(n) = 1/4.

a(n) = 8*A001651. - Danny Rorabaugh, Oct 22 2015

From Colin Barker, Aug 26 2016: (Start)

a(n) = 12*n-2*(-1)^n-6.

a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.

G.f.: 8*x*(1+x+x^2) / ((1-x)^2*(1+x)).

(End)

Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/72. - Amiram Eldar, Dec 31 2021

E.g.f.: 2*(4 + (6*x - 3)*exp(x) - exp(-x)). - David Lovler, Sep 05 2022

Extensions

Better name from Danny Rorabaugh, Oct 22 2015