A015468 q-Fibonacci numbers for q=10.

0, 1, 1, 11, 111, 11111, 1121111, 1112221111, 1122223221111, 11123333333221111, 112233445444433221111, 11123445566666555433221111, 1122345577889898877665433221111, 1112345679012233433220988765433221111

Offset

0,4

Links

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

Formula

a(n) = a(n-1) + 10^(n-2) a(n-2).

Maple

q:=10; 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]*10^(n-2)},  a, {n, 20}] (* 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, 10], {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)*(10^(n-2)): n in [1..15]]; // Vincenzo Librandi, Nov 09 2012
(PARI) q=10; 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, 10) for n in (0..20)] # G. C. Greubel, Dec 17 2019
(GAP) q:=10;; 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), this sequence (q=10), A015469 (q=11), A015470 (q=12).

Sequence in context: A261269 A031974 A117293 * A037842 A131293 A108047

Adjacent sequences: A015465 A015466 A015467 * A015469 A015470 A015471

Keyword

nonn,easy

Author

Olivier Gérard

Status

approved