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A192232
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Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)
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280
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1, 0, 2, 1, 6, 7, 22, 36, 89, 168, 377, 756, 1630, 3353, 7110, 14783, 31130, 65016, 136513, 285648, 599041, 1254456, 2629418, 5508097, 11542854, 24183271, 50674318, 106173180, 222470009, 466131960, 976694489, 2046447180, 4287928678, 8984443769, 18825088134
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OFFSET
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1,3
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COMMENTS
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Polynomial reduction: an introduction
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We begin with an example. Suppose that p(x) is a polynomial, so that p(x)=(x^2)t(x)+r(x) for some polynomials t(x) and r(x), where r(x) has degree 0 or 1. Replace x^2 by x+1 to get (x+1)t(x)+r(x), which is (x^2)u(x)+v(x) for some u(x) and v(x), where v(x) has degree 0 or 1. Continuing in this manner results in a fixed polynomial w(x) of degree 0 or 1. If p(x)=x^n, then w(x)=x*F(n)+F(n-1), where F=A000045, the sequence of Fibonacci numbers.
In order to generalize, write d(g) for the degree of an arbitrary polynomial g(x), and suppose that p, q, s are polynomials satisfying d(s)<d(q). By the division algorithm, there exists a unique pair t and r of polynomials such that p=q*t+r and d(r)<d(q). Replace q by s to get s*t+r, which is q*u+v for some u and v, where d(v)<d(q). Continue applying q->s in this manner until reaching w such that d(w)<d(q). We call w the reduction of p by q->s.
The coefficients of (reduction of p by q->s) comprise a vector of length d(q)-1, so that a sequence p(n,x) of polynomials begets a sequence of vectors, such as (F(n), F(n-1)) in the above example. We are interested in the component sequences (e.g., F(n-1) and F(n)) for various choices of p(n,x).
Following are examples of reduction by x^2->x+1:
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Suppose that b=(b(0), b(1),...) is a sequence, and let p(n,x)=b(0)+b(1)x+b(2)x^2+...+b(n)x^n. We define (reduction of sequence b by q->s) to be the vector given by (reduction of p(n,x) by q->s), with components in the order of powers, from 0 up to d(q)-1. For k=0,1,...,d(q)-1, we then have the "k-sequence of (reduction of sequence b by q->s)". Continuing the example, if b is the sequence given by b(k)=1 if k=n and b(k)=0 otherwise, then the 0-sequence of (reduction of b by x^2->x+1) is (F(n-1)), and the 1-sequence is (F(n)).
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For selected sequences b, here are the 0-sequences and 1-sequences of (reduction of b by x^2->x+1):
b=A000045, Fibonacci sequence (1,1,2,3,5,8,...) yields
b=(1,A000045)=(1,1,1,2,3,5,8,...) yields
b=A000027, natural number sequence (1,2,3,4,...) yields
b=A000032, Lucas sequence (1,3,4,7,11,...) yields
b=(A000217, triangular sequence (1,3,6,10,...) yields
b=(A000290, squares sequence (1,4,9,16,...) yields
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More comments:
(1) If s(n,x)=(reduction of x^n by q->s) and
p(x)=p(0)x^n+p(1)x^(n-1)+...+p(n)x^0, then
(reduction of p by q->s)=p(0)s(n,x)+p(1)s(n-1,x)
+...+p(n-1)s(1,x)+p(n)s(0,x). See A192744.
(2) For any polynomial p(x), let P(x)=(reduction of p(x)
by q->s). Then P(r)=p(r) for each zero r of
q(x)-s(x). In particular, if q(x)=x^2 and s(x)=x+1,
then P(r)=p(r) if r=(1+sqrt(5))/2 (golden ratio) or
r=(1-sqrt(5))/2.
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LINKS
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FORMULA
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Empirical G.f.: -x*(x^2+x-1)/(x^4+x^3-3*x^2-x+1). - Colin Barker, Sep 11 2012
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EXAMPLE
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The first four Fibonacci polynomials and their reductions by x^2->x+1 are shown here:
F1(x)=1 -> 1 + 0x
F2(x)=x -> 0 + 1x
F3(x)=x^2+1 -> 2+1x
F4(x)=x^3+2x -> 1+4x
F5(x)=x^4+3x^2+1 -> (x+1)^2+3(x+1)+1 -> 6+6x.
From these, read A192232=(1,0,1,1,6,...) and A112576=(0,1,1,4,6,...).
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MATHEMATICA
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q[x_] := x + 1;
reductionRules = {x^y_?EvenQ -> q[x]^(y/2), x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[FixedPoint[Expand[#1 /. reductionRules] &, Fibonacci[n, x]], {n, 1, 40}];
Table[Coefficient[Part[t, n], x, 0], {n, 1, 40}]
Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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