A192351 Coefficient of x in the reduction (by x^2->x+1) of polynomial p(n,x) identified in Comments.
0, 1, 2, 20, 56, 320, 1120, 5312, 20608, 90880, 368640, 1577984, 6522880, 27578368, 114909184, 483328000, 2020573184, 8480555008, 35502817280, 148874461184, 623609118720, 2614000353280, 10952269365248, 45901678641152, 192340840939520
Offset: 0
Keywords
Examples
The first four polynomials p(n,x) and their reductions are as follows: p(0,x)=1 -> 1 p(1,x)=x -> x p(2,x)=5+x+x^2 -> 6+2x p(3,x)=15x+3x^2+x^3 -> 4+20x. From these, we read A192350=(1,0,6,4,...) and A192351=(0,1,2,20...)
Links
- Robert Israel, Table of n, a(n) for n = 0..1606
Programs
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Maple
f:= gfun:-rectoproc({a(n) = 2*a(n-1)+12*a(n-2)-8*a(n-3)-16*a(n-4),a(0)=0,a(1)=1,a(2)=2,a(3)=20},a(n),remember): map(f, [$0..50]); # Robert Israel, Jan 01 2018 -
Mathematica
(See A192350.)
Formula
Conjecture: a(n) = 2*a(n-1)+12*a(n-2)-8*a(n-3)-16*a(n-4). G.f.: x*(4*x^2+1) / (16*x^4+8*x^3-12*x^2-2*x+1). [Colin Barker, Jan 17 2013]
Confirmation of conjecture by Robert Israel, Jan 01 2018: (Start)
The polynomials p(n,x) have g.f. G(z) = (1-x*z)/(1-2*x*z-5*z^2-x*z^2+x^2*z^2).
The reductions mod x^2-x-1 have g.f. g(z) = (1+x*z-2*z-6*z^2+4*x*z^3)/(1-2*z-12*z^2+8*z^3+16*z^4):
note that the numerator of G(z)-g(z) is divisible by x^2-x-1. (End)
Extensions
Offset corrected by Robert Israel, Jan 01 2018
Comments