A015459 -id:A015459 - cp's OEIS frontend

A015470 q-Fibonacci numbers for q=12, scaling a(n-2).

0, 1, 1, 13, 157, 22621, 3278173, 5632106845, 9794204234077, 201818365309759837, 4211530365904119214429, 1041342647528423104910537053, 260767900948768868884822059725149, 773726564635922870118341112574642827613

Offset: 0

Olivier Gérard

Vincenzo Librandi, Table of n, a(n) for n = 0..60

q-Fibonacci numbers: A000045 (q=1), A015459 (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), this sequence (q=12).

q:=12;; a:=[0,1];; for n in [3..20] do a[n]:=a[n-1]+q^(n-3)*a[n-2]; od; a; # G. C. Greubel, Dec 17 2019

[0] cat[n le 2 select 1 else Self(n-1) + Self(n-2)*(12^(n-2)): n in [1..15]]; // Vincenzo Librandi, Nov 09 2012

q:=12; 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 17 2019

RecurrenceTable[{a[0]==0, a[1]==1, a[n]==a[n-1]+a[n-2]*12^(n-2)},  a, {n, 60}] (* Vincenzo Librandi, Nov 09 2012 *)
F[n_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^(j^2), {j, 0, Floor[(n-1)/2]}];
Table[F[n, 12], {n, 0, 20}] (* G. C. Greubel, Dec 17 2019 *)

q=12; 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 17 2019

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,12) for n in (0..20)] # G. C. Greubel, Dec 17 2019

a(n) = a(n-1) + 12^(n-2)*a(n-2).

A280294 a(n) = a(n-1) + 2^n * a(n-2) with a(0) = 1 and a(1) = 1.

Original entry on oeis.org

1, 1, 5, 13, 93, 509, 6461, 71613, 1725629, 38391485, 1805435581, 80431196861, 7475495336637, 666367860021949, 123144883455482557, 21958686920654707389, 8092381769059159562941, 2886261393833112966453949, 2124255587862077437434059453

Offset: 0

Seiichi Manyama, Dec 31 2016

nonn

The Rogers-Ramanujan continued fraction is defined by R(q) = q^(1/5)/(1+q/(1+q^2/(1+q^3/(1+ ... )))). The limit of a(n)/A015459(n+2) is 2^(-1/5) * R(2).

			1/1 = a(0)/A015459(2).
1/(1+2/1) = 1/3 = a(1)/A015459(3).
1/(1+2/(1+2^2/1)) = 5/7 = a(2)/A015459(4).
1/(1+2/(1+2^2/(1+2^3/1))) = 13/31 = a(3)/A015459(5).

Seiichi Manyama, Table of n, a(n) for n = 0..114
Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction

Cf. A015463, A015459, A128915.

Cf. similar sequences with the recurrence a(n-1) + q^n * a(n-2) for n>1, a(0)=1 and a(1)=1: this sequence (q=2), A279543 (q=3), A280340 (q=10).

nxt[{n_,a_,b_}]:={n+1,b,b+2^(n+1)*a}; NestList[nxt,{1,1,1},20][[All,2]] (* Harvey P. Dale, Jul 17 2020 *)

def a():
    a, b, p = 1, 0, 1
    while True:
        p, a, b = p + p, b, b + p * a
        yield b
A280294 = a()
print([next(A280294) for  in range(19)]) # _Peter Luschny, Dec 05 2017

A165901 a(0)=0, a(1)=1, a(n) = a(n-1) + 2^(n-3)*a(n-2).

Original entry on oeis.org

0, 1, 1, 2, 4, 12, 44, 236, 1644, 16748, 227180, 4514668, 120830828, 4743850860, 252205386604, 19683018509164, 2085749545569132, 324572324799712108, 68670413434009029484, 21339842291507941739372, 9022108271913939454266220

Offset: 0

Jaume Oliver Lafont, Sep 29 2009

nonn

The difference equation y(n, x, s, q) = x*y(n-1, x, s, q) + q^(n-3)*y(n-2, x, s, q) yields the two-variable q-Fibonacci polynomials in the form y(n, x, s, q) = Sum_{j=0..floor((n-1)/2)} q-binomial(n-j-1,j, q)*q^(j*(j-1))*x^(n-2*j-1)*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 A015459 and A015473. - G. C. Greubel, Dec 17 2019

G. C. Greubel, Table of n, a(n) for n = 0..100

Cf. A015459.

q-Fibonacci numbers: A000045 (q=1), this sequence (q=2), A165902 (q=3).

q:=2;; a:=[0,1];; for n in [3..20] do a[n]:=a[n-1]+q^(n-3)*a[n-2]; od; a; # G. C. Greubel, Dec 19 2019

q:=2; I:=[0,1]; [n le 2 select I[n] else Self(n-1) + q^(n-3)*Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 19 2019

q:=2; seq(add((product((1-q^(n-j-1-k))/(1-q^(k+1)), k=0..j-1))* q^(j*(j-1)), j = 0..floor((n-1)/2)), n = 0..20); # G. C. Greubel, Dec 19 2019

F[n_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^(j*(j-1)), {j,0,Floor[(n-1)/2]}];
Table[F[n, 2], {n,0,20}] (* G. C. Greubel, Dec 19 2019 *)
nxt[{n_,a_,b_}]:={n+1,b,b+a*2^(n-2)}; NestList[nxt,{1,0,1},20][[All,2]] (* Harvey P. Dale, Oct 03 2021 *)

a(n) = if(n<2, n, a(n-1) + 2^(n-3)*a(n-2));

def F(n,q): return sum( q_binomial(n-j-1, j, q)*q^(j*(j-1)) for j in (0..floor((n-1)/2)))
[F(n,2) for n in (0..20)] # G. C. Greubel, Dec 19 2019

A136680 Triangle T(n, k) = f(k) for k < n+1, otherwise 0, where f(k) = f(k-1) + k^(k-2)*f(k-2) with f(0) = 0 and f(1) = 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 4, 20, 1, 1, 4, 20, 520, 1, 1, 4, 20, 520, 26440, 1, 1, 4, 20, 520, 26440, 8766080, 1, 1, 4, 20, 520, 26440, 8766080, 6939853440, 1, 1, 4, 20, 520, 26440, 8766080, 6939853440, 41934828744960, 1, 1, 4, 20, 520, 26440, 8766080, 6939853440, 41934828744960, 694027278828744960

Offset: 1

Roger L. Bagula, Apr 06 2008

			Triangle begins as:
  1;
  1, 1;
  1, 1, 4;
  1, 1, 4, 20;
  1, 1, 4, 20, 520;
  1, 1, 4, 20, 520, 26440;
  1, 1, 4, 20, 520, 26440, 8766080;
  1, 1, 4, 20, 520, 26440, 8766080, 6939853440;
  1, 1, 4, 20, 520, 26440, 8766080, 6939853440, 41934828744960;

G. C. Greubel, Rows n = 1..30 of the triangle, flattened

q-Fibonacci numbers include: A015459, A015460, A015461, A015462, A015463, A015464, A015465, A015467, A015468, A015469, A015470, A015473, A015474, A015475, A015476, A015477, A015479, A015480, A015481, A015482, A015484, A015485.

function f(k)
  if k lt 2 then return k;
  else return f(k-1) + k^(k-2)*f(k-2);
  end if; return f;
end function;
A136680:= func< n,k | k le n select f(k) else 0 >;
[A136680(n,k): k in [1..n], n in [1..14]]; // G. C. Greubel, Dec 01 2022

T[k_]:= T[k]= If[k<2, k, T[k-1] + n^(k-2)*T[k-2]];
Table[T[k], {n,10}, {k,n}]//Flatten

@CachedFunction
def f(k):
    if (k<2): return k
    else: return f(k-1) + k^(k-2)*f(k-2)
def A136680(n,k): return f(k) if (k < n+1) else 0
flatten([[A136680(n,k) for k in range(1,n+1)] for n in range(1,15)]) # G. C. Greubel, Dec 01 2022

T(k) = T(k-1) + n^(k-2)*T(k-2), with T(0) = 0, T(1) = 1.

T(n, k) = f(k) for k < n+1, otherwise 0, where f(k) = f(k-1) + k^(k-2)*f(k-2) with f(0) = 0 and f(1) = 1. - G. C. Greubel, Dec 01 2022

Edited by G. C. Greubel, Dec 01 2022

A241497 q-Pell numbers with q=2.

Original entry on oeis.org

0, 1, 2, 6, 20, 88, 496, 3808, 39360, 566144, 11208448, 312282624, 12102016000, 663758845952, 50897375227904, 5539307216494592, 844981210166968320, 183201981290428727296, 55743092552083293274112

Offset: 0

Ralf Stephan, Apr 24 2014

nonn

Cf. A015459.

# sage -i ore_algebra
from ore_algebra import *
R. = QQ['x']; A. = OreAlgebra(R, 'Qx', q=2)
print((Qx^2 - 2*Qx - x).to_list([0,1], 10))

Recurrence: a(n) = 2*a(n-1) + 2^(n-2) a(n-2), a(0) = 0, a(1) = 1.

A241498 q-Lucas numbers with q=2.

Original entry on oeis.org

2, 1, 3, 5, 17, 57, 329, 2153, 23209, 298793, 6240297, 159222313, 6549286441, 332636583465, 27158513845801, 2752117405591081, 447717208255194665, 90629100354663736873, 29432224060567101302313, 11908369665747052420720169, 7727389313799049256214259241, 6251142704628989668810750223913

Offset: 0

Ralf Stephan, Apr 24 2014

nonn

a(n) = 2k+1, where apparently k = 8m, m odd for n > 3.

More generally, a(k) is congruent to a(n) modulo 2^(n-1) for any k > n. - Charlie Neder, Mar 09 2019

Hao Pan, Congruences for q-Lucas Numbers, Electron. J. Combin., 20, Issue 2 (2013), P29.

Cf. A015459, A241497.

# sage -i ore_algebra
from ore_algebra import *
R. = QQ['x']; A. = OreAlgebra(R, 'Qx', q=2)
print((Qx^2 - Qx - x).to_list([2,1], 10))

Recurrence: a(n) = a(n-1) + 2^(n-2)*a(n-2), starting 2, 1.

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A015470 q-Fibonacci numbers for q=12, scaling a(n-2).

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A280294 a(n) = a(n-1) + 2^n * a(n-2) with a(0) = 1 and a(1) = 1.

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A165901 a(0)=0, a(1)=1, a(n) = a(n-1) + 2^(n-3)*a(n-2).

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A136680 Triangle T(n, k) = f(k) for k < n+1, otherwise 0, where f(k) = f(k-1) + k^(k-2)*f(k-2) with f(0) = 0 and f(1) = 1, read by rows.

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A241497 q-Pell numbers with q=2.

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A241498 q-Lucas numbers with q=2.

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