A047584 -id:A047584 - cp's OEIS frontend

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

José María Grau Ribas, Jul 12 2015

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

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).

Cf. A000914.

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

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 *)

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

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)

Better name from Danny Rorabaugh, Oct 22 2015

A047418 Numbers that are congruent to {0, 2, 3, 4, 6} mod 8.

Original entry on oeis.org

0, 2, 3, 4, 6, 8, 10, 11, 12, 14, 16, 18, 19, 20, 22, 24, 26, 27, 28, 30, 32, 34, 35, 36, 38, 40, 42, 43, 44, 46, 48, 50, 51, 52, 54, 56, 58, 59, 60, 62, 64, 66, 67, 68, 70, 72, 74, 75, 76, 78, 80, 82, 83, 84, 86, 88, 90, 91, 92, 94, 96, 98, 99, 100, 102

Offset: 1

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Cf. A047412, A047434, A047486, A047489, A047501, A047511, A047584.

Filtered([0..103],n->n mod 8 = 0 or n mod 8 = 2 or n mod 8 = 3 or n mod 8 = 4 or n mod 8 = 6); # Muniru A Asiru, Oct 23 2018

[n : n in [0..150] | n mod 8 in [0, 2, 3, 4, 6]]; // Wesley Ivan Hurt, Aug 08 2016

A047418:=n->8*floor(n/5)+[(0, 2, 3, 4, 6)][(n mod 5)+1]: seq(A047418(n), n=0..100); # Wesley Ivan Hurt, Aug 08 2016

Select[Range[0,100], MemberQ[{0,2,3,4,6}, Mod[#,8]]&] (* or *) LinearRecurrence[{1,0,0,0,1,-1}, {0,2,3,4,6,8}, 70] (* Harvey P. Dale, Oct 01 2015 *)

G.f.: x^2*(2 + x + x^2 + 2*x^3 + 2*x^4)/((x^4 + x^3 + x^2 + x + 1)*(x - 1)^2). - R. J. Mathar, Dec 05 2011
From Wesley Ivan Hurt, Aug 08 2016: (Start)
a(n) = a(n-1) + a(n-5) - a(n-6) for n > 6, a(n) = a(n-5) + 8 for n > 5.
a(n) = (40*n - 45 - 2*(n mod 5) + 3*((n + 1) mod 5) + 3*((n + 2) mod 5) - 2*((n + 3) mod 5) - 2*((n + 4) mod 5))/25.
a(5*k) = 8*k - 2, a(5*k-1) = 8*k - 4, a(5*k-2) = 8*k - 5, a(5*k-3) = 8*k - 6, a(5*k-4) = 8*k - 8. (End)
a(n) = (40*n - 45 + 2*cos(2*Pi*(n - 1)/5) - 2*cos(2*Pi*n/5) -  2*cos(4*Pi*n/5) - 6*cos(2*Pi*(n + 1)/5) - 6*cos(Pi*(2*n + 1)/5) + 6*cos(2*Pi*(2*n + 1)/5) - 2*cos(Pi*(4*n + 1)/5) + 6*sin(Pi*(8*n + 3)/10))/25. - Wesley Ivan Hurt, Oct 10 2018

cp's OEIS Frontend

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

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