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A192882
Coefficient of x in the reduction by (x^2 -> x+1) of the polynomial p(n,x) given in Comments.
3
0, 1, 3, 14, 51, 205, 792, 3107, 12117, 47362, 184965, 722591, 2822544, 11025793, 43069611, 168242270, 657200859, 2567211037, 10028243016, 39173122739, 153021167805, 597743469778, 2334953116653, 9120979734623, 35629097057568
OFFSET
0,3
COMMENTS
The polynomial p(n,x) is defined by p(0,x) = 1, p(1,x) = x, and p(n,x) = 2*x*p(n-1,x) + (x^2)*p(n-1,x). See A192872.
FORMULA
a(n) = 2*a(n-1) + 7*a(n-2) + 2*a(n-3) - a(n-4).
G.f.: x*(1+x+x^2) / ( 1-2*x-7*x^2-2*x^3+x^4 ). - R. J. Mathar, May 07 2014
a(n) = Fibonacci(n)*Pell-Lucas(n)/2, where Pell-Lucas(n) = A002203(n). - G. C. Greubel, Jul 29 2019
MATHEMATICA
(* First program *)
q = x^2; s = x + 1; z = 25;
p[0, x_]:= 1; p[1, x_]:= x;
p[n_, x_]:= 2 p[n-1, x]*x + p[n-2, x]*x^2;
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}] (* A192880 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192882 *)
FindLinearRecurrence[u1]
FindLinearRecurrence[u2]
(* Additional programs *)
LinearRecurrence[{2, 7, 2, -1}, {0, 1, 3, 14}, 30] (* G. C. Greubel, Jan 08 2019 *)
Table[Fibonacci[n]*LucasL[n, 2]/2, {n, 0, 30}] (* G. C. Greubel, Jul 29 2019 *)
PROG
(PARI) my(x='x+O('x^30)); concat([0], Vec(x*(1+x+x^2)/(1-2*x-7*x^2-2*x^3 +x^4))) \\ G. C. Greubel, Jan 08 2019
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!( x*(1+x+x^2)/(1-2*x-7*x^2-2*x^3+x^4) )); // G. C. Greubel, Jan 08 2019
(Sage) (x*(1+x+x^2)/(1-2*x-7*x^2-2*x^3+x^4)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 08 2019
(GAP) a:=[0, 1, 3, 14];; for n in [5..30] do a[n]:=2*a[n-1]+7*a[n-2] +2*a[n-3] -a[n-4]; od; a; # G. C. Greubel, Jan 08 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jul 11 2011
STATUS
approved