OFFSET
0,2
COMMENTS
If the reduction (x^2 + c)^n by x^3 -> x^2 + c is applied to the polynomials (x^2+c)^n for c=1 instead of c=2, the results are as follows:
A052554: constant terms,
A052529: coefficients of x,
A124820: coefficients of x^2.
Those three sequences satisfy the recurrence:
u(n) = 4*u(n-1) - 3*u(n-2) + u(n-3).
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-12,8).
FORMULA
a(n) = 7*a(n-1) - 12*a(n-2) + 8*a(n-3).
G.f.: (1-x)*(1-4*x)/(1-7*x+12*x^2-8*x^3). - Colin Barker, Jul 26 2012
MATHEMATICA
q = x^3; s = x^2 + 2; z = 40;
p[n_, x_] := (x^2 + 2)^n;
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}] (* A192808 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192809 *)
u3 = Table[Coefficient[Part[t, n], x, 2], {n, 1, z}] (* A192810 *)
uu = u2/2, {n, 1, z}] (* A192811 *)
LinearRecurrence[{7, -12, 8}, {1, 2, 6}, 50] (* G. C. Greubel, Jan 02 2019 *)
PROG
(PARI) my(x='x+O('x^30)); Vec((1-x)*(1-4*x)/(1-7*x+12*x^2-8*x^3)) \\ G. C. Greubel, Jan 02 2019
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)*(1-4*x)/(1-7*x+12*x^2-8*x^3) )); // G. C. Greubel, Jan 02 2019
(Sage) ((1-x)*(1-4*x)/(1-7*x+12*x^2-8*x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 02 2019
(GAP) a:=[1, 2, 6];; for n in [4..25] do a[n]:=7*a[n-1]-12*a[n-2]+8*a[n-3]; od; Print(a); # Muniru A Asiru, Jan 02 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 10 2011
STATUS
approved