A015469 q-Fibonacci numbers for q=11.

0, 1, 1, 12, 133, 16105, 1963358, 2595689713, 3480804151551, 50586130104323474, 746191869036731097905, 119280194867984161366496439, 19354414621214347335584253057344, 34032051023004810891710239239325511573

Offset

0,4

Links

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

Formula

a(n) = a(n-1) + 11^(n-2)*a(n-2).

Maple

q:=11; 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

Mathematica

RecurrenceTable[{a[0]==0, a[1]==1, a[n]==a[n-1]+a[n-2]*11^(n-2)},  a, {n, 61}] (* 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, 11], {n, 0, 20}] (* G. C. Greubel, Dec 17 2019 *)

Prog

(Magma) [0] cat[n le 2 select 1 else Self(n-1) + Self(n-2)*(11^(n-2)): n in [1..15]]; // Vincenzo Librandi, Nov 09 2012
(PARI) q=11; 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
(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, 11) for n in (0..20)] # G. C. Greubel, Dec 17 2019
(GAP) q:=11;; 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

Crossrefs

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), this sequence (q=11), A015470 (q=12).

Sequence in context: A016123 A353148 A015457 * A144785 A214994 A208440

Adjacent sequences: A015466 A015467 A015468 * A015470 A015471 A015472

Keyword

nonn,easy

Author

Olivier Gérard

Status

approved