A176476 Partial sums of A012814.
0, 1, 6, 27, 113, 464, 1896, 7738, 31571, 128800, 525455, 2143647, 8745216, 35676948, 145547524, 593775045, 2422362078, 9882257735, 40315615409, 164471408184, 670976837020, 2737314167774, 11167134898975, 45557394660800, 185855747875875, 758216295635151
Offset: 0
Examples
a(5) = A000931(2) + A000931(7) + A000931(12) + A000931(17) + A000931(22) + A000931(27) = 0 + 1 + 5 + 21 + 86 + 351 = 464.
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..1638
- Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 7.
- Index entries for linear recurrences with constant coefficients, signature (6,-9,5,-1).
Programs
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Mathematica
LinearRecurrence[{6,-9,5,-1},{0,1,6,27},30] (* Harvey P. Dale, Feb 08 2025 *)
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PARI
a(n) = my(v=vector(n+1), u=[0,1,6,27]); for(k=1, n+1, v[k]=if(k<=4, u[k], 5*v[k-1] - 4*v[k-2] + v[k-3] + 1)); v[n+1] \\ Jianing Song, Feb 04 2019
Formula
a(n) = A012855(n+3) - 1. a(n) = 6*a(n-1) - 9*a(n-2) + 5*a(n-3) - a(n-4). - R. J. Mathar, Oct 18 2010
G.f.: x/(1 - 6*x + 9*x^2 - 5*x^3 + x^4). - Colin Barker, Feb 03 2012
From Jianing Song, Feb 04 2019: (Start)
a(n+3) = 5*a(n+2) - 4*a(n+1) + a(n) + 1.
Extensions
New name, more terms and a(0) = 0 prepended by Jianing Song, Feb 04 2019
Comments