cp's OEIS Frontend

Polynomial reduction: an introduction

...

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)s in this manner until reaching w such that d(w)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:

n-th Fibonacci p(x) -> A192232+x*A112576

n-th cyclotomic p(x) -> A192233+x*A051258

n-th 1st-kind Chebyshev p(x) -> A192234+x*A071101

n-th 2nd-kind Chebyshev p(x) -> A192235+x*A192236

x(x+1)(x+2)...(x+n-1) -> A192238+x*A192239

(x+1)^n -> A001519+x*A001906

(x^2+x+1)^n -> A154626+x*A087635

(x+2)^n -> A020876+x*A030191

(x+3)^n -> A192240+x*A099453

...

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)).

...

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

0-sequence A166536 and 1-sequence A064831.

b=(1,A000045)=(1,1,1,2,3,5,8,...) yields

0-sequence A166516 and 1-sequence A001654.

b=A000027, natural number sequence (1,2,3,4,...) yields

0-sequence A190062 and 1-sequence A122491.

b=A000032, Lucas sequence (1,3,4,7,11,...) yields

0-sequence A192243 and 1-sequence A192068.

b=A000217, triangular sequence (1,3,6,10,...) yields

0-sequence A192244 and 1-sequence A192245.

b=A000290, squares sequence (1,4,9,16,...) yields

0-sequence A192254 and 1-sequence A192255.

More examples: A192245-A192257.

...

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.

This constant and its reciprocal are the real solutions of x^4 - 2*x^3 - 2*x^2 - 2*x + 1 = (x^2 - (sqrt(5)+1)*x + 1)*(x^2 + (sqrt(5)-1)*x + 1) = 0.

This constant and its reciprocal are the solutions of x^2 - (1+sqrt(5))*x + 1 = 0.

Decimal expansion of the largest x satisfying x^2 - (1+sqrt(5))*x + 1 = 0.

For sequences of type aa(n) = 2*(aa(n-1) + aa(n-2) + aa(n-3)) - aa(n-4) for arbitrary initial terms (except the trivial all zero), i.e., linear recurrence relations of order 4 with signature (2,2,2,-1), lim_{n -> infinity} aa(n)/aa(n-1) = this constant; see for instance A192234, A192237, A317973, A317974, A317975, A317976.