A100302 Expansion of (1 - x - 6*x^2)/((1 - x)*(1 - x - 8*x^2)).
1, 1, 3, 5, 23, 57, 235, 685, 2559, 8033, 28499, 92757, 320743, 1062793, 3628731, 12131069, 41160911, 138209457, 467496739, 1573172389, 5313146295, 17898525401, 60403695755, 203591898957, 686821464991, 2315556656641, 7810128376563, 26334581629685, 88815608642183
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (2,7,-8).
Programs
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Magma
I:=[1,1,3]; [n le 3 select I[n] else 2*Self(n-1) +7*Self(n-2) -8*Self(n-3): n in [1..31]]; // G. C. Greubel, Feb 04 2023
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Mathematica
LinearRecurrence[{2,7,-8},{1,1,3},29] (* Stefano Spezia, Sep 08 2022 *)
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SageMath
def A100302(n): return (3 + lucas_number1(n+1, 1, -8))/4 [A100302(n) for n in range(31)] # G. C. Greubel, Feb 04 2023
Formula
a(n) = 2*a(n-1) + 7*a(n-2) - 8*a(n-3).
a(n) = 3/4 + (1/(4*sqrt(33)))*(((1 + sqrt(33))/2)^(n+1) - ((1 - sqrt(33))/2)^(n+1)).
E.g.f.: 3*exp(x)/4 + exp(x/2)*(33*cosh(sqrt(33)*x/2) + sqrt(33)*sinh(sqrt(33)*x/2))/132. - Stefano Spezia, Sep 08 2022
a(n) = (1/4)*(3 + A015443(n+1)). - G. C. Greubel, Feb 04 2023
Comments