OFFSET
0,3
COMMENTS
The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x + 1, and p(n,x) = x*p(n-1,x) + 2*(x^2)*p(n-1,x) + 1. See A192872.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,6,-5,-6,4).
FORMULA
a(n) = 2*a(n-1) + 6*a(n-2) - 5*a(n-3) - 6*a(n-4) + 4*a(n-5).
G.f.: (1-x-4*x^2) / ( (1-x)*(1+x-x^2)*(1-2*x-4*x^2) ). - R. J. Mathar, May 06 2014
MAPLE
seq(coeff(series((1-x-4*x^2)/((1-x)*(1+x-x^2)*(1-2*x-4*x^2)), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Jan 08 2019
MATHEMATICA
q = x^2; s = x + 1; z = 26;
p[0, x_] := 1; p[1, x_] := x + 1;
p[n_, x_] := p[n - 1, x]*x + 2*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}];
u0 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192876 *)
u1 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192877 *)
FindLinearRecurrence[u0]
FindLinearRecurrence[u1]
LinearRecurrence[{2, 6, -5, -6, 4}, {1, 1, 4, 9, 31}, 26] (* Ray Chandler, Aug 02 2015 *)
PROG
(PARI) my(x='x+O('x^30)); Vec((1-x-4*x^2)/((1-x)*(1+x-x^2)*(1-2*x-4*x^2) )) \\ G. C. Greubel, Jan 08 2019
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x-4*x^2)/((1-x)*(1+x-x^2)*(1-2*x-4*x^2)) )); // G. C. Greubel, Jan 08 2019
(Sage) ((1-x-4*x^2)/((1-x)*(1+x-x^2)*(1-2*x-4*x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jan 08 2019
(GAP) a:=[1, 1, 4, 9, 31];; for n in [6..30] do a[n]:=2*a[n-1]+6*a[n-2] - 5*a[n-3]-6*a[n-4]+4*a[n-5]; od; a; # G. C. Greubel, Jan 08 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jul 11 2011
STATUS
approved