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A259754
Numbers that are congruent to {3,9,15,18,21} mod 24.
6
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
1,1
COMMENTS
Original name: Numbers n such that n/A259748(n) = 3/2.
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)
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 *)
PROG
(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
CROSSREFS
Cf. A000914.
Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259755.
Sequence in context: A043381 A100331 A155764 * A277569 A310329 A310330
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
Better name from Danny Rorabaugh, Oct 22 2015
STATUS
approved