A192386 -id:A192386 - cp's OEIS frontend

A192387 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.

0, 2, 4, 32, 96, 512, 1856, 8576, 33792, 147456, 602112, 2566144, 10637312, 44892160, 187269120, 787087360, 3292069888, 13812760576, 57837355008, 242497880064, 1015868817408, 4258009186304, 17841063460864, 74771320537088, 313317428035584

Offset: 1

Clark Kimberling, Jun 30 2011

The polynomial p(n,x) is defined by ((x+d)^n - (x-d)^n)/(2*d), where d = sqrt(x+5). For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.

			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 A192386 = (1, 0, 8, 8, 96, ...) and a(n) = (0, 2, 4, 32, 96, ...).

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).

Cf. A112576, A192232, A192386, A192388.

R:=PowerSeriesRing(Integers(), 41);
[0] cat Coefficients(R!( 2*x^2/(1-2*x-12*x^2+8*x^3+16*x^4) )); // G. C. Greubel, Jul 10 2023

(See A192386.)
LinearRecurrence[{2,12,-8,-16}, {0,2,4,32}, 40] (* G. C. Greubel, Jul 10 2023 *)

@CachedFunction
def a(n): # a = A192387
    if (n<5): return (0,0,2,4,32)[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

From Colin Barker, Dec 09 2012: (Start)
a(n) = 2*a(n-1) + 12*a(n-2) - 8*a(n-3) - 16*a(n-4).
G.f.: 2*x^2/(1-2*x-12*x^2+8*x^3+16*x^4). (End)
a(n) = 2^n*A112576(n). - R. J. Mathar, Mar 08 2021
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) = Sum_{k=0..n-1} T(n, k)*Fibonacci(k). (End)

A192388 a(n) = A192387(n)/2.

Original entry on oeis.org

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

Clark Kimberling, Jun 30 2011

			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, ...).

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).

Cf. A112576, A192232, A192386, A192387.

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

(* See A192386 *)
LinearRecurrence[{2,12,-8,-16}, {0,1,2,16}, 40] (* G. C. Greubel, Jul 10 2023 *)

@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

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)

cp's OEIS Frontend

A192387 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.

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