

A192351


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


2



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
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OFFSET

0,3


COMMENTS

To define the polynomials p(n,x), let d=sqrt(x+5); then p(n,x)=(1/2)((x+d)^n+(xd)^n). These are similar to polynomials at A161516.
For an introduction to reductions of polynomials by substitutions such as x^2>x+1, see A192232.


LINKS

Robert Israel, Table of n, a(n) for n = 0..1606


FORMULA

Conjecture: a(n) = 2*a(n1)+12*a(n2)8*a(n3)16*a(n4). G.f.: x*(4*x^2+1) / (16*x^4+8*x^312*x^22*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) = (1x*z)/(12*x*z5*z^2x*z^2+x^2*z^2).
The reductions mod x^2x1 have g.f. g(z) = (1+x*z2*z6*z^2+4*x*z^3)/(12*z12*z^2+8*z^3+16*z^4):
note that the numerator of G(z)g(z) is divisible by x^2x1. (End)


EXAMPLE

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...)


MAPLE

f:= gfun:rectoproc({a(n) = 2*a(n1)+12*a(n2)8*a(n3)16*a(n4), 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.)


CROSSREFS

Cf. A192232, A192350.
Sequence in context: A139271 A133217 A001504 * A136905 A230644 A183907
Adjacent sequences: A192348 A192349 A192350 * A192352 A192353 A192354


KEYWORD

nonn


AUTHOR

Clark Kimberling, Jun 28 2011


EXTENSIONS

Offset corrected by Robert Israel, Jan 01 2018


STATUS

approved



