OFFSET
0,3
COMMENTS
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 + 1.
LINKS
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).
FORMULA
a(n) = 3*a(n-1) + a(n-3) + a(n-4).
G.f.: (1-2*x-x^2)/(1-3*x-x^3-x^4). - Colin Barker, Aug 31 2012
MATHEMATICA
(* To obtain very general results, delete the next line. *)
u = 1; v = 1; a = 1; b = 1; c = 1; 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}] (* p(0, x), p(1, x), ... p(5, x) *)
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, 1, 2, 7}, 30] (* G. C. Greubel, Jan 11 2019 *)
PROG
(PARI) my(x='x+O('x^30)); Vec((1-2*x-x^2)/(1-3*x-x^3-x^4)) \\ G. C. Greubel, Jan 11 2019
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-2*x-x^2)/(1-3*x-x^3-x^4) )); // G. C. Greubel, Jan 11 2019
(Sage) ((1-2*x-x^2)/(1-3*x-x^3-x^4)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 11 2019
(GAP) a:=[1, 1, 2, 7];; for n in [5..30] do a[n]:=3*a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 11 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 12 2011
STATUS
approved