A291223 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^4.
0, 0, 1, 1, 3, 5, 8, 17, 25, 52, 83, 159, 271, 497, 868, 1572, 2762, 4984, 8784, 15799, 27939, 50089, 88831, 158880, 282293, 504179, 896780, 1600335, 2848339, 5080363, 9045953, 16129172, 28726972, 51209648, 91223508, 162594868, 289675121, 516264093
Offset: 0
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (0, 4, 1, -5, -1, 4, 0, -1)
Programs
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Mathematica
z = 60; s = x/(1 - x^2); p = 1 - s^3 - s^4; Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *) Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291223 *)
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PARI
concat(vector(2), Vec(x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8) + O(x^50))) \\ Colin Barker, Aug 25 2017
Formula
a(n) = 4*a(n-2) + a(n-3) - 5*a(n-4) - a(n-5) + 4*a(n-6) - a(n-8) for n >= 9.
G.f.: x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8). - Colin Barker, Aug 25 2017
Comments