A192744 -id:A192744 - cp's OEIS frontend

A192232 Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)

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

Offset: 1

Clark Kimberling, Jun 26 2011

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.

			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,...).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-1).

Cf. A168561, A192233 - A192240, A192744.

Cf. A206296, A265398, A265399, A265752, A265753.

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}]
  (* A192232 *)
Table[Coefficient[Part[t, n], x, 1], {n, 1, 40}]
(* A112576 *)
(* Peter J. C. Moses, Jun 25 2011 *)
LinearRecurrence[{1, 3, -1, -1}, {1, 0, 2, 1}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

Vec((1-x-x^2)/(1-x-3*x^2+x^3+x^4)+O(x^99)) \\ Charles R Greathouse IV, Jan 08 2013

Empirical G.f.: -x*(x^2+x-1)/(x^4+x^3-3*x^2-x+1). - Colin Barker, Sep 11 2012
The above formula is correct. - Charles R Greathouse IV, Jan 08 2013
a(n) = A265752(A206296(n)). - Antti Karttunen, Dec 15 2015
a(n) = A112576(n) -A112576(n-1) -A112576(n-2). - R. J. Mathar, Dec 16 2015

Example corrected by Clark Kimberling, Dec 18 2017

A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

Original entry on oeis.org

0, 1, 3, 9, 20, 40, 74, 131, 225, 379, 630, 1038, 1700, 2773, 4511, 7325, 11880, 19252, 31182, 50487, 81725, 132271, 214058, 346394, 560520, 906985, 1467579, 2374641, 3842300, 6217024, 10059410, 16276523, 26336025, 42612643, 68948766

Offset: 0

Clark Kimberling, Jul 13 2011

The titular polynomials are defined recursively:  p(n,x) = x*p(n-1,x) + 3n - 1, with p(0,x)=1.  For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232 and A192744.
...
The list of examples at A192744 is extended here; the recurrence is given by p(n,x) = x*p(n-1,x) + v(n), with p(0,x)=1, and the reduction of p(n,x) by x^2 -> x+1 is represented by u1 + u2*x:
...
If v(n)=        n, then u1=A001595, u2=A104161.
If v(n)=      n-1, then u1=A001610, u2=A066982.
If v(n)=     3n-1, then u1=A171516, u2=A192951.
If v(n)=     3n-2, then u1=A192746, u2=A192952.
If v(n)=     2n-1, then u1=A111314, u2=A192953.
If v(n)=      n^2, then u1=A192954, u2=A192955.
If v(n)=   -1+n^2, then u1=A192956, u2=A192957.
If v(n)=    1+n^2, then u1=A192953, u2=A192389.
If v(n)=   -2+n^2, then u1=A192958, u2=A192959.
If v(n)=    2+n^2, then u1=A192960, u2=A192961.
If v(n)=    n+n^2, then u1=A192962, u2=A192963.
If v(n)=   -n+n^2, then u1=A192964, u2=A192965.
If v(n)= n(n+1)/2, then u1=A030119, u2=A192966.
If v(n)= n(n-1)/2, then u1=A192967, u2=A192968.
If v(n)= n(n+3)/2, then u1=A192969, u2=A192970.
If v(n)=     2n^2, then u1=A192971, u2=A192972.
If v(n)=   1+2n^2, then u1=A192973, u2=A192974.
If v(n)=  -1+2n^2, then u1=A192975, u2=A192976.
If v(n)=  1+n+n^2, then u1=A027181, u2=A192978.
If v(n)=  1-n+n^2, then u1=A192979, u2=A192980.
If v(n)=  (n+1)^2, then u1=A001891, u2=A053808.
If v(n)=  (n-1)^2, then u1=A192981, u2=A192982.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (3, -2, -1, 1).

Cf. A000045, A192232, A192744.

F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2)-(3*n+5)); # G. C. Greubel, Jul 12 2019

I:=[0, 1, 3, 9]; [n le 4 select I[n] else 3*Self(n-1)-2*Self(n-2)-1*Self(n-3)+Self(n-4): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

F:=Fibonacci; [F(n+4)+2*F(n+2)-(3*n+5): n in [0..40]]; // G. C. Greubel, Jul 12 2019

(* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + 3n - 1;
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}] (* A171516 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192951 *)
(* Additional programs *)
LinearRecurrence[{3,-2,-1,1},{0,1,3,9},40] (* Vincenzo Librandi, Nov 16 2011 *)
With[{F=Fibonacci}, Table[F[n+4]+2*F[n+2]-(3*n+5), {n,0,40}]] (* G. C. Greubel, Jul 12 2019 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,-1,-2,3]^n*[0;1;3;9])[1,1] \\ Charles R Greathouse IV, Mar 22 2016

vector(40, n, n--; f=fibonacci; f(n+4)+2*f(n+2)-(3*n+5)) \\ G. C. Greubel, Jul 12 2019

f=fibonacci; [f(n+4)+2*f(n+2)-(3*n+5) for n in (0..40)] # G. C. Greubel, Jul 12 2019

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).
From Bruno Berselli, Nov 16 2011: (Start)
G.f.: x*(1+2*x^2)/((1-x)^2*(1 - x - x^2)).
a(n) = ((25+13*t)*(1+t)^n + (25-13*t)*(1-t)^n)/(10*2^n) - 3*n - 5 = A000285(n+2) - 3*n - 5  where t=sqrt(5). (End)
a(n) = Fibonacci(n+4) + 2*Fibonacci(n+2) - (3*n+5). - G. C. Greubel, Jul 12 2019

A192872 Constant term in the reduction by (x^2 -> x+1) of the polynomial p(n,x) given in Comments.

Original entry on oeis.org

1, 0, 3, 4, 13, 30, 81, 208, 547, 1428, 3741, 9790, 25633, 67104, 175683, 459940, 1204141, 3152478, 8253297, 21607408, 56568931, 148099380, 387729213, 1015088254, 2657535553, 6957518400

Offset: 0

Clark Kimberling, Jul 11 2011

The polynomial p(n,x) is defined by p(0,x)=1, p(1,x)=x, and p(n,x) = x*p(n-1,x) + (x^2)*p(n-1,x) + 1. The resulting sequence typifies a general class which we shall describe here. Suppose that u,v,a,b,c,d,e,f are numbers used to define these polynomials:
...
q(x) = x^2
s(x) = u*x + v
p(0,x) = a, p(1,x) = b*x + c
p(n,x) = d*x*p(n-1,x) + e*(x^2)*p(n-2,x) + f.
...
We shall assume that u is not 0 and that {d,e} is not {0}.  The reduction of p(n,x) by the repeated substitution q(x)->s(x), as defined and described at A192232 and A192744, has the form h(n)+k(n)*x.  The numerical sequences h and k are, formally, linear recurrence sequences of order 5.  The second Mathematica program below shows initial terms and the recurrence coefficients, which are too long to be included here, which imply these properties:
(1) The numbers a,b,c,f affect initial terms but not the recurrence coefficients, which depend only on u,v,d,e.
(2) If v=0 or e=0, the order of recurrence is <= 3.
(3) If v=0 and e=0, the order of recurrence is 2, and the coefficients are 1+d*u and d*u.
(See A192904 for similar results for other p(n,x).)
...
Examples:
u v a b c d e f seq h.....seq k
1 1 1 2 0 1 1 0 -A121646..A059929
1 1 1 3 0 1 1 0 A128533...A081714
1 1 2 1 0 1 1 0 A081714...A001906
1 1 1 1 1 1 1 0 A000045...A001906
1 1 2 1 1 1 1 0 A129905...A192879
1 1 1 2 1 1 1 0 A061646...A079472
1 1 1 1 0 1 1 1 A192872...A192873
1 1 1 1 1 2 1 1 A192874...A192875
1 1 1 1 1 2 1 1 A192876...A192877
1 1 1 1 1 1 2 1 A192880...A192882
1 1 1 1 1 1 1 1 A166536...A064831
The terms of several of these sequences are products of Fibonacci numbers (A000045), or Fibonacci numbers and Lucas numbers (A000032).

			The coefficients in all the polynomials p(n,x) are Fibonacci numbers (A000045).  The first six and their reductions:
p(0,x) = 1 -> 1
p(1,x) = x -> x
p(2,x) = 1 + 2*x^2 -> 3 + 2*x
p(3,x) = 1 + x + 3*x^3 -> 4 + 7*x
p(4,x) = 1 + x + 2*x^2 + 5*x^4 -> 13 + 18*x
p(5,x) = 1 + x + 2*x^2 + 3*x^3 + 8*x^5 -> 30 + 49*x

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A192232, A192744, A192873, A192908 (sums of adjacent terms).

a:=[1,0,3,4];; for n in [5..30] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 06 2019

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1)) )); // G. C. Greubel, Jan 06 2019

(* First program *)
q = x^2; s = x + 1; z = 26;
p[0, x_] := 1; p[1, x_] := x;
p[n_, x_] := p[n - 1, x]*x + p[n - 2, x]*x^2 + 1;
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}] (* A192872 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192873 *)
(* End of 1st program *)
(* ******************************************** *)
(* Second program: much more general *)
(* u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 1; Nine degrees of freedom for user; shown values generate A192872. *)
q = x^2; s = u*x + v; z = 11;
(* will apply reduction (x^2 -> u*x+v) to p(n,x) *)
p[0, x_] := a; p[1, x_] := b*x + c;
(* initial values of polynomial sequence p(n,x) *)
p[n_, x_] := d*x*p[n - 1, x] + e*(x^2)*p[n - 2, x] + f;
(* recurrence for p(n,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}];
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}];
Simplify[FindLinearRecurrence[u1]] (* for 0-sequence *)
Simplify[FindLinearRecurrence[u2]] (* for 1-sequence *)
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, 4}]
(* initial values for 0-sequence *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, 4}]
(* initial values for 1-sequence *)
LinearRecurrence[{3,0,-3,1},{1,0,3,4},26] (* Ray Chandler, Aug 02 2015 *)

my(x='x+O('x^30)); Vec((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))) \\ G. C. Greubel, Jan 06 2019

((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 06 2019

a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).

G.f.: (2*x-1)*(x^2-x+1) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - R. J. Mathar, Oct 26 2011

A192616 Constant term in the reduction of the n-th Fibonacci polynomial by x^3->x^2+x+1. See Comments.

Original entry on oeis.org

1, 0, 1, 1, 2, 6, 10, 29, 57, 142, 309, 720, 1625, 3714, 8457, 19259, 43902, 99970, 227830, 518943, 1182401, 2693624, 6136837, 13980960, 31851853, 72565704, 165320833, 376638417, 858066430, 1954869262, 4453630790, 10146374277, 23115721705

Offset: 1

Clark Kimberling, Jul 09 2011

For discussions of polynomial reduction, see A192232 and A192744.

			The first five polynomials p(n,x) and their reductions:
F1(x)=1 -> 1
F2(x)=x -> x
F3(x)=x^2+1 -> x^2+1
F4(x)=x^3+2x -> x^2+3x+1
F5(x)=x^4+3x^2+1 -> 4x^2+2x+2, so that
A192616=(1,0,1,1,2,...), A192617=(0,1,0,3,2,...), A192651=(0,0,1,1,5,...)

Index entries for linear recurrences with constant coefficients, signature (1,4,-1,-4,1,1).

Cf. A192744, A192232, A192617, A192651.

q = x^3; s = x^2 + x + 1; z = 40;
p[n_, x_] := Fibonacci[n, x];
Table[Expand[p[n, x]], {n, 1, 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, 1, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
  (* A192616 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192617 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}]
  (* A192651 *)

a(n) = a(n-1)+4*a(n-2)-a(n-3)-4a(n-4)+a(n-5)+a(n-6).

G.f.: -x*(x^4+x^3-3*x^2-x+1)/(x^6+x^5-4*x^4-x^3+4*x^2+x-1). [Colin Barker, Jul 27 2012]

A192904 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) defined below at Comments.

Original entry on oeis.org

1, 0, 1, 5, 16, 49, 153, 480, 1505, 4717, 14784, 46337, 145233, 455200, 1426721, 4471733, 14015632, 43928817, 137684905, 431542080, 1352570689, 4239325789, 13287204352, 41645725825, 130529073953, 409113752000, 1282274186177

Offset: 0

Clark Kimberling, Jul 12 2011

The titular polynomial is defined by p(n,x) = (x^2)*p(n-1,x) + x*p(n-2,x), with p(0,x) = 1, p(1,x) = x.  The resulting sequence typifies a general class which we shall describe here. Suppose that u,v,a,b,c,d,e,f are numbers used to define these polynomials:
...
q(x) = x^2
s(x) = u*x + v
p(0,x) = a, p(1,x) = b*x + c
p(n,x) = d*(x^2)*p(n-1,x) + e*x*p(n-2,x) + f.
...
We shall assume that u is not 0 and that {d,e} is not {0}.  The reduction of p(n,x) by the repeated substitution q(x) -> s(x), as defined and described at A192232 and A192744, has the form h(n) + k(n)*x.  The numerical sequences h and k are linear recurrence sequences, formally of order 5.  The Mathematica program below, with first line deleted, shows initial terms and recurrence coefficients, which imply these properties:
(1)  the recurrence coefficients depend only on u,v,d,e; the parameters a,b,c,f affect only the initial terms.
(2)  if e=0 or v=0, the order of recurrence is <= 3;
(3)  if e=0 and v=0, the recurrence coefficients are 1+d*u^2 and -d*u^2 (cf. similar results at A192872).
...
Examples:
u v a b c d e f... seq h.....seq k
1 1 1 1 1 1 0 0... A001906..A001519
1 1 1 1 0 0 1 0... A103609..A193609
1 1 1 1 0 1 1 0... A192904..A192905
1 1 1 1 1 1 0 0... A001519..A001906
1 1 1 1 1 1 1 0... A192907..A192907
1 1 1 1 1 1 0 1... A192908..A069403
1 1 1 1 1 1 1 1... A192909..A192910
The terms of these sequences involve Fibonacci numbers, F(n)=A000045(n); e.g.,
A001906: even-indexed F(n)
A001519: odd-indexed F(n)
A103609: (1,1,1,1,2,2,3,3,5,5,8,8,...)

			The first six polynomials and reductions:
1 -> 1
x -> x
x + x^3 -> 1 + 3*x
x^2 + x^3 + x^5 -> 5 + 8*x
x^2 + 2*x^4 + x^5 + x^7 -> 16 + 25*x
x^3 + 2*x^4 + 3*x^6 + x^7 + x^9 -> 49 + 79*x, so that
A192904 = (1,0,1,5,16,49,...) and
A192905 = (0,1,3,8,25,79,...)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,1,1).

Cf. A192232, A192744, A192905, A192872.

a:=[1,0,1,5];; for n in [5..40] do a[n]:=3*a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 10 2019

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4) )); // G. C. Greubel, Jan 10 2019

(* To obtain general results, delete the next line. *)
u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 0;
q = x^2; s = u*x + v; z = 24;
p[0, x_] := a; p[1, x_] := b*x + c;
p[n_, x_] :=  d*(x^2)*p[n - 1, x] + e*x*p[n - 2, x] + f;
Table[Expand[p[n, x]], {n, 0, 8}]
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}];
u0 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192904 *)
u1 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192905 *)
Simplify[FindLinearRecurrence[u0]] (* recurrence for 0-sequence *)
Simplify[FindLinearRecurrence[u1]] (* recurrence for 1-sequence *)
LinearRecurrence[{3,0,1,1}, {1,0,1,5}, 40] (* G. C. Greubel, Jan 10 2019 *)

my(x='x+O('x^40)); Vec((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)) \\ G. C. Greubel, Jan 10 2019

((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 10 2019

a(n) = 3*a(n-1) + a(n-3) + a(n-4).

G.f.: (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4). - Colin Barker, Aug 31 2012

A192772 Constant term in the reduction of the n-th Fibonacci polynomial by x^3->x^2+2x+1.

Original entry on oeis.org

1, 0, 1, 1, 2, 7, 12, 41, 86, 247, 585, 1548, 3849, 9896, 25001, 63724, 161721, 411257, 1044878, 2655719, 6748972, 17151849, 43589578, 110777391, 281529169, 715471992, 1818293377, 4620978640, 11743694657, 29845241080, 75848270001

Offset: 1

Clark Kimberling, Jul 09 2011

For discussions of polynomial reduction, see A192232 and A192744.

			The first five polynomials p(n,x) and their reductions are as follows:
F1(x)=1 -> 1
F2(x)=x -> x
F3(x)=x^2+1 -> x^2+1
F4(x)=x^3+2x -> x^2+4x+1
F5(x)=x^4+3x^2+1 -> 6x^2+3x+2, so that
A192772=(1,0,1,1,2,...), A192773=(0,1,0,4,3,...), A192774=(0,0,1,1,6,...)

Index entries for linear recurrences with constant coefficients, signature (1,5,-1,-5,1,1).

Cf. A192744, A192232, A192616, A192773, A192774.

q = x^3; s = x^2 + 2 x + 1; z = 40;
p[n_, x_] := Fibonacci[n, x];
Table[Expand[p[n, x]], {n, 1, 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, 1, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192772 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192773 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}] (* A192774 *)

a(n) = a(n-1)+5*a(n-2)-a(n-3)-5*a(n-4)+a(n-5)+a(n-6).

G.f.: -x*(x^2-x-1)*(x^2+2*x-1) / (x^6+x^5-5*x^4-x^3+5*x^2+x-1). [Colin Barker, Jan 17 2013]

A192777 Constant term in the reduction of the n-th Fibonacci polynomial by x^3->x^2+3x+1. See Comments.

Original entry on oeis.org

1, 0, 1, 1, 2, 8, 14, 55, 121, 392, 989, 2912, 7797, 22104, 60553, 169289, 467622, 1300888, 3603914, 10008543, 27755249, 77034176, 213702153, 593005504, 1645265209, 4565154816, 12666317073, 35144684065, 97512548090, 270561677224

Offset: 1

Clark Kimberling, Jul 09 2011

nonn

For discussions of polynomial reduction, see A192232 and A192744.

			The first five polynomials p(n,x) and their reductions are as follows:
F1(x)=1 -> 1
F2(x)=x -> x
F3(x)=x^2+1 -> x^2+1
F4(x)=x^3+2x -> x^2+5x+1
F5(x)=x^4+3x^2+1 -> 7x^2+4x+2, so that
A192777=(1,0,1,1,2,...), A192778=(0,1,0,5,4,...), A192779=(0,0,1,1,7,...)

Index entries for linear recurrences with constant coefficients, signature (1,6,-1,-6,1,1).

Cf. A192744, A192232, A192616, A192772, A192778, A192779.

q = x^3; s = x^2 + 3 x + 1; z = 40;
p[n_, x_] := Fibonacci[n, x];
Table[Expand[p[n, x]], {n, 1, 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, 1, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
  (* A192777 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192778 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}]
  (* A192779 *)

a(n)=a(n-1)+6*a(n-2)-a(n-3)-6*a(n-4)+a(n-5)+a(n-6).

G.f.: -x*(1-5*x^2+x^4-x+x^3) / ( (x^2-x-1)*(x^4+2*x^3-3*x^2-2*x+1) ). - R. J. Mathar, May 06 2014

A192778 Coefficient of x in the reduction of the n-th Fibonacci polynomial by x^3->x^2+3x+1.

Original entry on oeis.org

0, 1, 0, 5, 4, 28, 48, 183, 424, 1315, 3420, 9864, 26756, 75237, 207128, 577345, 1597624, 4439764, 12307388, 34166643, 94769936, 262998791, 729644824, 2024614928, 5617339496, 15586328073, 43245649904, 119991232893, 332929027020

Offset: 1

Clark Kimberling, Jul 09 2011

For discussions of polynomial reduction, see A192232 and A192744.

			The first five polynomials p(n,x) and their reductions:
F1(x)=1 -> 1
F2(x)=x -> x
F3(x)=x^2+1 -> x^2+1
F4(x)=x^3+2x -> x^2+5x+1
F5(x)=x^4+3x^2+1 -> 7x^2+4x+2, so that
A192777=(1,0,1,1,2,...), A192778=(0,1,0,5,4,...), A192779=(0,0,1,1,7,...)

Index entries for linear recurrences with constant coefficients, signature (1,6,-1,-6,1,1).

Cf. A192744, A192232, A192616, A192772, A192777, A192779.

a(n) = a(n-1)+6*a(n-2)-a(n-3)-6*a(n-4)+a(n-5)+a(n-6).

G.f.: x^2*(x^2+x-1)/((x^2-x-1)*(x^4+2*x^3-3*x^2-2*x+1)). [Colin Barker, Nov 23 2012]

A192779 Coefficient of x^2 in the reduction of the n-th Fibonacci polynomial by x^3->x^2+3x+1.

Original entry on oeis.org

0, 0, 1, 1, 7, 12, 47, 107, 337, 868, 2520, 6808, 19192, 52756, 147185, 407069, 1131599, 3136292, 8707655, 24151335, 67025633, 185946904, 515971328, 1431563056, 3972149312, 11021051864, 30579529249, 84846231017, 235416993159, 653192251196

Offset: 1

Clark Kimberling, Jul 09 2011

For discussions of polynomial reduction, see A192232 and A192744.

			The first five polynomials p(n,x) and their reductions:
F1(x)=1 -> 1
F2(x)=x -> x
F3(x)=x^2+1 -> x^2+1
F4(x)=x^3+2x -> x^2+5x+1
F5(x)=x^4+3x^2+1 -> 7x^2+4x+2, so that
A192777=(1,0,1,1,2,...), A192778=(0,1,0,5,4,...), A192779=(0,0,1,1,7,...)

Index entries for linear recurrences with constant coefficients, signature (1,6,-1,-6,1,1).

Cf. A192744, A192232, A192616, A192772, A192777.

(See A192777.)
LinearRecurrence[{1,6,-1,-6,1,1},{0,0,1,1,7,12}, 30] (* Harvey P. Dale, Oct 29 2018 *)

a(n) = a(n-1)+6*a(n-2)-a(n-3)-6*a(n-4)+a(n-5)+a(n-6).

G.f.: -x^3/((x^2-x-1)*(x^4+2*x^3-3*x^2-2*x+1)). [Colin Barker, Nov 23 2012]

A192798 Constant term in the reduction of the n-th Fibonacci polynomial by x^3->x^2+2. See Comments.

Original entry on oeis.org

1, 0, 1, 2, 3, 10, 17, 42, 87, 188, 411, 876, 1907, 4100, 8863, 19134, 41289, 89174, 192459, 415542, 897049, 1936576, 4180809, 9025544, 19484825, 42064320, 90809993, 196043706, 423225563, 913674090, 1972469945, 4258235410, 9192822255

Offset: 1

Clark Kimberling, Jul 10 2011

For discussions of polynomial reduction, see A192232 and A192744.

			The first five polynomials p(n,x) and their reductions:
F1(x)=1 -> 1
F2(x)=x -> x
F3(x)=x^2+1 -> x^2+1
F4(x)=x^3+2x -> x^2+2x+2
F5(x)=x^4+3x^2+1 -> 4x^2+2*x+3, so that
A192798=(1,0,1,2,3,...), A192799=(0,1,0,2,2,...), A192800=(0,0,1,1,4,...)

Index entries for linear recurrences with constant coefficients, signature (1,3,0,-3,1,1).

Cf. A192744, A192232, A192616, A192799, A192800.

q = x^3; s = x^2 + 2; z = 40;
p[n_, x_] := Fibonacci[n, x];
Table[Expand[p[n, x]], {n, 1, 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, 1, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]  (* A192798 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192799 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}]
  (* A192800 *)

a(n) = a(n-1)+3*a(n-2)-3*a(n-4)+a(n-5)+a(n-6).

G.f.: -x*(x-1)*(x+1)*(x^2+x-1)/(x^6+x^5-3*x^4+3*x^2+x-1). [Colin Barker, Jul 27 2012]

Comment in Mathematica code corrected by Colin Barker, Jul 27 2012

cp's OEIS Frontend

A192232 Constant term of the reduction of n-th Fibonacci polynomial by x^2 -> x+1. (See Comments.)

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A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

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A192872 Constant term in the reduction by (x^2 -> x+1) of the polynomial p(n,x) given in Comments.

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A192616 Constant term in the reduction of the n-th Fibonacci polynomial by x^3->x^2+x+1. See Comments.

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A192904 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) defined below at Comments.

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A192772 Constant term in the reduction of the n-th Fibonacci polynomial by x^3->x^2+2x+1.

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