A192388 a(n) = A192387(n)/2.
0, 1, 2, 16, 48, 256, 928, 4288, 16896, 73728, 301056, 1283072, 5318656, 22446080, 93634560, 393543680, 1646034944, 6906380288, 28918677504, 121248940032, 507934408704, 2129004593152, 8920531730432, 37385660268544, 156658714017792
Offset: 1
Examples
The first five polynomials p(n,x) and their reductions are as follows: p(0,x) = 1 -> 1 p(1,x) = 2*x -> 2*x p(2,x) = 3 + x + 3*x^2 -> 8 + 4*x p(3,x) = 12*x + 4*x^2 + 4*x^3 -> 8 + 32*x p(4,x) = 9 + 6*x + 31*x^2 + 10*x^3 + 5*x^4 -> 96 + 96*x. From these, read a(n) = (0, 2, 4, 32, 96, ...)/2 = (0, 1, 2, 16, 48, ...).
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (2,12,-8,-16).
Programs
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Magma
R
:=PowerSeriesRing(Integers(), 41); [0] cat Coefficients(R!( x^2/(1-2*x-12*x^2+8*x^3+16*x^4) )); // G. C. Greubel, Jul 10 2023 -
Mathematica
(* See A192386 *) LinearRecurrence[{2,12,-8,-16}, {0,1,2,16}, 40] (* G. C. Greubel, Jul 10 2023 *)
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SageMath
@CachedFunction def a(n): # a = A192388 if (n<5): return (0,0,1,2,16)[n] else: return 2*a(n-1) +12*a(n-2) -8*a(n-3) -16*a(n-4) [a(n) for n in range(1,41)] # G. C. Greubel, Jul 10 2023
Formula
From G. C. Greubel, Jul 10 2023: (Start)
T(n, k) = [x^k] ((x+sqrt(x+5))^n - (x-sqrt(x+5))^n)/(2*sqrt(x+5)).
a(n) = (1/2)*Sum_{k=0..n-1} T(n, k)*Fibonacci(k-1).
a(n) = 2*a(n-1) + 12*a(n-2) - 8*a(n-3) - 16*a(n-4).
G.f.: x^2/(1-2*x-12*x^2+8*x^3+16*x^4).
a(n) = 2^(n-1)*A112576(n). (End)