A192944 - OEIS

%I #8 Jul 27 2019 19:29:40

%S 0,1,4,19,127,1227,17977,407108,14510651,821907178,74498246381,

%T 10849935064552,2545826568211757,963950723522943935,

%U 589590299737652176495,582892188767255266969095,931834379222684656945128850,2409387593714287957763063565225

%N Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x)=(x+1)(x+2)...(x+F(n+1)), where F=A000045, the Fibonacci sequence.

%C For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.

%H G. C. Greubel, <a href="/A192944/b192944.txt">Table of n, a(n) for n = 0..97</a>

%e The first four polynomials p(n,x) and their reductions are as follows:

%e p(0,x) = 1

%e p(1,x) = x+1 -> 1+x

%e p(2,x) = (x+1)(x+2) -> 3+4x

%e p(3,x) = (x+1)(x+2)(x+3) -> 13+19x

%e From these, read

%e A192943=(1,1,3,13,...) and A192944=(0,1,4,19,...)

%t q = x^2; s = x + 1; z = 26;

%t p[0, x]:= 1;

%t p[n_, x_]:= (x + Fibonacci[n+1])*p[n-1, x];

%t Table[Expand[p[n, x]], {n, 0, 7}]

%t reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]

%t t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];

%t u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192943 *)

%t u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192944 *)

%Y Cf. A192232, A192744, A192943.

%K nonn

%O 0,3

%A _Clark Kimberling_, Jul 13 2011