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A192427
Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.
2
0, 1, 1, 8, 11, 45, 80, 251, 517, 1432, 3195, 8317, 19360, 48827, 116213, 288360, 694331, 1708397, 4138480, 10138363, 24636645, 60217912, 146570491, 357833309, 871703360, 2126857275, 5183425493, 12642971912, 30819571387, 75160150861
OFFSET
0,4
COMMENTS
The polynomial p(n,x) is defined by ((x+d)/2)^n + ((x-d)/2)^n, where d = sqrt(x^2+8). For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.
FORMULA
From Colin Barker, May 12 2014: (Start)
a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 4*a(n-4).
G.f.: x*(1+2*x^2)/(1-x-5*x^2+2*x^3+4*x^4). (End)
a(n) = Sum_{k=0..n} T(n, k)*Fibonacci(k), where T(n, k) = [x^k] ((x + sqrt(x^2+8))^n + (x - sqrt(x^2+8))^n)/2^n. - G. C. Greubel, Jul 13 2023
EXAMPLE
The first five polynomials p(n,x) and their reductions are as follows:
p(0,x) = 2 -> 2
p(1,x) = x -> x
p(2,x) = 4 + x^2 -> 5 + x
p(3,x) = 6*x + x^3 -> 1 + 8*x
p(4,x) = 8 + 8*x^2 + x^4 -> 18 + 11*x.
From these, read A192426 = (2, 0, 5, 1, 18, ...) and a(n) = (0, 1, 1, 8, 11, ...).
MATHEMATICA
(See A192426.)
LinearRecurrence[{1, 5, -2, -4}, {0, 1, 1, 8}, 40] (* G. C. Greubel, Jul 13 2023 *)
PROG
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+2*x^2)/(1-x-5*x^2+2*x^3+4*x^4) )); // G. C. Greubel, Jul 13 2023
(SageMath)
@CachedFunction
def a(n): # a = A192427
if (n<4): return (0, 1, 1, 8)[n]
else: return a(n-1) + 5*a(n-2) - 2*a(n-3) - 4*a(n-4)
[a(n) for n in range(41)] # G. C. Greubel, Jul 13 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 30 2011
EXTENSIONS
Typo in name corrected by G. C. Greubel, Jul 13 2023
STATUS
approved