A330795 Evaluation of the polynomials given by the Riordan square of the Fibonacci sequence with a(0) = 1 (A193737) at 1/2 and normalized with 2^n.

1, 3, 9, 39, 153, 615, 2457, 9831, 39321, 157287, 629145, 2516583, 10066329, 40265319, 161061273, 644245095, 2576980377, 10307921511, 41231686041, 164926744167, 659706976665, 2638827906663, 10555311626649, 42221246506599, 168884986026393, 675539944105575

Offset: 0

Author

Peter Luschny, Jan 10 2020

Keywords

• nonn
• easy

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,4).

Crossrefs

Cf. A006131, A015521, A193737, A321620, A324969 (Fibonacci with a(0)=1).

Programs

Magma

[1] cat [3*(4^n -(-1)^n)/5: n in [1..30]]; // G. C. Greubel, Sep 14 2023

Maple

gf := (4*x^2 - 1)/(x*(4*x + 3) - 1): ser := series(gf, x, 32):
seq(coeff(ser, x, n), n=0.. 25);
# Alternative:
gf:= (3/5)*exp(-x)*(exp(5*x) - 1) + 1: ser := series(gf, x, 32):
seq(n!*coeff(ser, x, n), n=0.. 25);
# Or:
a := proc(n) option remember; if n < 3 then return [1, 3, 9][n + 1] fi;
4*a(n-2) + 3*a(n-1) end: seq(a(n), n=0..25);

Mathematica

LinearRecurrence[{3,4}, {1,3,9}, 31] (* G. C. Greubel, Sep 14 2023 *)

SageMath

[3*(4^n -(-1)^n)//5 + int(n==0) for n in range(31)] # G. C. Greubel, Sep 14 2023

Formula

a(n) = 2^n*Sum_{k=0..n} A193737(n,k)/2^k.

a(n) = [x^n] (1 - 4*x^2)/(1 - x*(3 + 4*x)).

a(n) = n! [x^n] (3/5)*exp(-x)*(exp(5*x) - 1) + 1.

a(n) = 4*a(n-2) + 3*a(n-1).

a(n) = 3*A015521(n), n>0. - R. J. Mathar, Aug 19 2022