OFFSET
0,3
COMMENTS
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
LINKS
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).
FORMULA
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
EXAMPLE
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
MATHEMATICA
(* 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 *)
PROG
(PARI) 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
(Magma) m:=30; R<x>:=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
(Sage) ((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
(GAP) 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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 11 2011
STATUS
approved