A015469 q-Fibonacci numbers for q=11, scaling a(n-2).
0, 1, 1, 12, 133, 16105, 1963358, 2595689713, 3480804151551, 50586130104323474, 746191869036731097905, 119280194867984161366496439, 19354414621214347335584253057344, 34032051023004810891710239239325511573
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..60
Crossrefs
Programs
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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
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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
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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
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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 *)
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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
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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
Formula
a(n) = a(n-1) + 11^(n-2)*a(n-2).