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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.

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

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

p(0,x) = 1

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

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

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

(See A192943.)

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

p[0, x]:= 1;

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

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

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

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

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

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

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_Clark Kimberling (ck6(AT)evansville.edu), _, Jul 13 2011

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allocated for Clark KimberlingCoefficient 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.

0, 1, 4, 19, 127, 1227, 17977, 407108, 14510651, 821907178, 74498246381, 10849935064552, 2545826568211757, 963950723522943935, 589590299737652176495, 582892188767255266969095, 931834379222684656945128850, 2409387593714287957763063565225

0,3

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

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

p(0,x)=1

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

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

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

From these, read

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

(See A192943.)

Cf. A192232, A192744, A192943.

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Clark Kimberling (ck6(AT)evansville.edu), Jul 13 2011

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allocated for Clark Kimberling

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