A015459 q-Fibonacci numbers for q=2, scaling a(n-2).

0, 1, 1, 3, 7, 31, 143, 1135, 10287, 155567, 2789039, 82439343, 2938415279, 171774189743, 12207523172527, 1419381685547183, 201427441344229551, 46711726513354322095, 13247460522448782176431, 6135846878080826487812271

Offset: 0

Author

Olivier Gérard

Keywords

• nonn
• easy

Comments

From Gary W. Adamson, Apr 17 2009: (Start)

Preface the series with another "1": (1, 1, 1, 3, 7, ..., a(n)).

Then a(n+2) = (1, 1, 1, 3, 7, ..., a(n)) dot (1, 2, 4, 8, ...).

Example: (143) = (1, 1, 1, 3, 7) dot (1, 2, 4, 8, 16) = (1 + 2 + 4 + 24 + 112).

Analogous procedures apply to other q-Fibonacci sequences for q(n). (End)

The difference equation y(n, x, s) = x*y(n-1, x, s) + q^(n-2)*s*y(n-2, x, s) yields a type of two variable q-Fibonacci polynomials in the form F(n, x, s, q) = Sum_{j=0..floor((n-1)/2)} q-binomial(n-j-1,j, q)*q^(j^2)*x^(n-2*j)*s^j. When x=s=1 these polynomials reduce to q-Fibonacci numbers. This family of q-Fibonacci numbers is different from that of the q-Fibonacci numbers defined in A015473. - G. C. Greubel, Dec 17 2019

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..100.
• L. Carlitz, Fibonacci notes 3: q-Fibonacci Numbers, Fibonacci Quarterly 12 (1974), pp. 317-322.
• Eric Weisstein's World of Mathematics, q-Analog.

Crossrefs

q-Fibonacci numbers: A000045 (q=1), this sequence (q=2), A015460 (q=3), A015461 (q=4), A015462 (q=5), A015463 (q=6), A015464 (q=7), A015465 (q=8), A015467 (q=9), A015468 (q=10), A015469 (q=11), A015470 (q=12).

Differs from A015473.

Programs

GAP

q:=2;; a:=[0,1];; for n in [3..30] do a[n]:=a[n-1]+q^(n-3)*a[n-2]; od; a; # G. C. Greubel, Dec 16 2019

Magma

[0] cat [n le 2 select 1 else Self(n-1) + Self(n-2)*(2^(n-2)): n in [1..20]]; // Vincenzo Librandi, Nov 08 2012

Maple

q:=2; seq(add((product((1-q^(n-j-1-k))/(1-q^(k+1)), k=0..j-1))*q^(j^2), j = 0..floor((n-1)/2)), n = 0..20); # G. C. Greubel, Dec 16 2019

Mathematica

RecurrenceTable[{a[0]==0, a[1]==1, a[n]==a[n-1]+a[n-2]*2^(n-2)}, a, {n, 20}] (* Vincenzo Librandi, Nov 08 2012 *)
F[n_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^(j^2), {j, 0, Floor[(n-1)/2]}];
Table[F[n, 2], {n, 0, 20}] (* G. C. Greubel, Dec 16 2019 *)

PARI

q=2; m=20; v=concat([0,1], vector(m-2)); for(n=3, m, v[n]=v[n-1]+q^(n-3)*v[n-2]); v \\ G. C. Greubel, Dec 16 2019

Python

def a():
    a, b, p = 0, 1, 1
    while True:
        yield a
        p, a, b = p + p, b, b + p * a
A015463 = a()
print([next(A015463) for  in range(20)]) # _Peter Luschny, Dec 05 2017

Sage

from ore_algebra import *
R. = QQ['x']
A. = OreAlgebra(R, 'Qx', q=2)
print((Qx^2 - Qx - x).to_list([0,1], 10))  # Ralf Stephan, Apr 24 2014

Sage

def F(n,q): return sum( q_binomial(n-j-1, j, q)*q^(j^2) for j in (0..floor((n-1)/2)))
[F(n,2) for n in (0..20)] # G. C. Greubel, Dec 16 2019

Formula

a(n) = a(n-1) + 2^(n-2)*a(n-2).

Associated constant: C_2 = lim_{n->oo} a(2*n)*a(2*n-2)/a(2*n-1)^2 = 1.225306147422043724739386133... (C_1=1). - Benoit Cloitre, Aug 30 2003 [Formula corrected by Vaclav Kotesovec, Dec 05 2017]

a(n)*a(n+3) - a(n)*a(n+2) - 2*a(n+1)*a(n+2) + 2*a(n+1)^2 = 0. - Emanuele Munarini, Dec 05 2017

From Vaclav Kotesovec, Dec 05 2017: (Start)

a(n) ~ c * 2^(n*(n-2)/4), where

c = 2.815179026313038425026160599838001991828247939843695... if n is even and

c = 3.024413799639405763259604599843170276573526808693115... if n is odd. (End)