A259754 Numbers that are congruent to {3,9,15,18,21} mod 24.

3, 9, 15, 18, 21, 27, 33, 39, 42, 45, 51, 57, 63, 66, 69, 75, 81, 87, 90, 93, 99, 105, 111, 114, 117, 123, 129, 135, 138, 141, 147, 153, 159, 162, 165, 171, 177, 183, 186, 189, 195, 201, 207, 210, 213, 219, 225, 231, 234, 237, 243, 249, 255, 258, 261, 267

Offset: 1

Author

José María Grau Ribas, Jul 12 2015

Keywords

• nonn
• easy

Comments

Original name: Numbers n such that n/A259748(n) = 3/2.

Links

• Danny Rorabaugh, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Crossrefs

Cf. A000914.

Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259755.

Programs

Mathematica

A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 2/3 &]
Rest@ CoefficientList[Series[3 x (1 + x) (1 + x + x^2 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)), {x, 0, 56}], x] (* Michael De Vlieger, Aug 25 2016 *)
LinearRecurrence[{1,0,0,0,1,-1},{3,9,15,18,21,27},60] (* Harvey P. Dale, Aug 30 2016 *)

PARI

Vec(3*x*(1+x)*(1+x+x^2+x^4)/((1-x)^2*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

Formula

A259748(a(n))/a(n) = 2/3.

a(n) = 3*A047584(n). - Michel Marcus, Jul 18 2015

From Colin Barker, Aug 25 2016: (Start)

a(n) = a(n-1)+a(n-5)-a(n-6) for n>6.

G.f.: 3*x*(1+x)*(1+x+x^2+x^4) / ((1-x)^2*(1+x+x^2+x^3+x^4)).

(End)

Extensions

Better name from Danny Rorabaugh, Oct 22 2015