A027286 a(n) = Sum_{k=0..2n} (k+1) * A026584(n, k).
1, 4, 18, 56, 190, 564, 1722, 4976, 14454, 40940, 115698, 322728, 896558, 2471588, 6786090, 18537184, 50459366, 136844892, 370030434, 997705240, 2683514526, 7201203988, 19284880794, 51546789456, 137541880150, 366412976332
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,-16).
Programs
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Magma
I:=[1,4,18,56]; [n le 4 select I[n] else 2*Self(n-1) +7*Self(n-2) -8*Self(n-3) -16*Self(n-4): n in [1..31]]; // G. C. Greubel, Dec 12 2021
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Mathematica
LinearRecurrence[{2,7,-8,-16},{1,4,18,56}, 30] (* G. C. Greubel, Dec 12 2021 *) -
PARI
a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -16,-8,7,2]^n*[1;4;18;56])[1,1] \\ Charles R Greathouse IV, Oct 21 2022
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Sage
[2^(n-1)*(n+1)*(14*lucas_number2(n+2, 1/2, -1) + 5*lucas_number2(n+1, 1/2, -1))/17 for n in (0..30)] # G. C. Greubel, Dec 12 2021
Formula
G.f.: (1+2*x+3*x^2)/(1-x-4*x^2)^2.
From G. C. Greubel, Dec 12 2021: (Start)
a(n) = 2^(n-3)*( -6*F(n+1, 1/2) + Sum_{j=0..n} F(n-j+1, 1/2)*( 14*F(j+1, 1/2) + 5*F(j, 1/2) ), where F(n, x) are the Fibonacci polynomials.
a(n) = (2^(n-1)/17)*(n+1)*( 14*L(n+2, 1/2) + 5*L(n+1, 1/2) ), where L(n, x) are the Lucas polynomials.
a(n) = 2*a(n-1) + 7*a(n-2) - 8*a(n-3) - 16*a(n-4). (End)