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A192429
Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.
2
0, 1, 4, 21, 76, 329, 1256, 5157, 20216, 81505, 322924, 1293189, 5144644, 20550089, 81881168, 326756661, 1302722672, 5196774145, 20723304532, 82657204533, 329642305468, 1314745861769, 5243461810232, 20912613564549, 83404589311592
OFFSET
0,3
COMMENTS
Constant term of the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.
The polynomial p(n,x) is defined by ((x+d)^n + (x-d)^n)/2 + ((x+d)^n - (x-d)^n)/(2*d), where d = sqrt(x^2+4), as at A163762. For an introduction to reductions of polynomials by substitutions such as x^2 -> x+2, see A192232.
FORMULA
From Colin Barker, May 12 2014: (Start)
a(n) = 2*a(n-1) + 10*a(n-2) - 6*a(n-3) - 9*a(n-4).
G.f.: x*(1+2*x+3*x^2)/(1-2*x-10*x^2+6*x^3+9*x^4). (End)
a(n) = Sum_{k=0..n} T(n,k)*Fibonacci(k), where T(n, k) = [x^k] ( ((x + sqrt(x+4))^n + (x - sqrt(x+4))^n)/2 + ((x + sqrt(x+4))^n - (x - sqrt(x+4))^n)/(2*sqrt(x+4)) ). - G. C. Greubel, Jul 13 2023
EXAMPLE
The first five polynomials p(n,x) and their reductions are as follows:
p(0,x) = 1 -> 1
p(1,x) = 1 + x -> 1 + x
p(2,x) = 4 + 3*x + x^2 -> 5 + 4*x
p(3,x) = 4 + 13*x + 6*x^2 + x^3 -> 11 + 21*x
p(4,x) = 16 + 24*x + 29*x^2 + 10*x^3 + x^4 -> 57 + 76*x.
From these, read A192428 = (1, 1, 5, 11, 57, 185, ...) and a(n) = (0, 1, 4, 21, 76, 329, ...).
MATHEMATICA
(See A192428.)
LinearRecurrence[{2, 10, -6, -9}, {0, 1, 4, 21}, 40] (* G. C. Greubel, Jul 13 2023 *)
PROG
(Magma) R<x>:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+2*x+3*x^2)/(1-2*x-10*x^2+6*x^3+9*x^4) )); // G. C. Greubel, Jul 13 2023
(SageMath)
@CachedFunction
def a(n): # a = A192429
if (n<4): return (0, 1, 4, 21)[n]
else: return 2*a(n-1) + 10*a(n-2) - 6*a(n-3) - 9*a(n-4)
[a(n) for n in range(41)] # G. C. Greubel, Jul 13 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jun 30 2011
STATUS
approved