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A015461
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q-Fibonacci numbers for q=4.
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13
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0, 1, 1, 5, 21, 341, 5717, 354901, 23771733, 5838469717, 1563742763605, 1532083548256853, 1641235215638133333, 6427665390003549698645, 27541785384957544314239573, 431380864280640133787922528853
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OFFSET
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0,4
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LINKS
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FORMULA
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a(n) = a(n-1) + 4^(n-2)*a(n-2).
Associated constant: C_4 = lim_{n->infinity} a(n)*a(n-2)/a(n-1)^2 = 1.094337777197221121533242886... . - Benoit Cloitre, Aug 30 2003
a(n)*a(n+3) - a(n)*a(n+2) - 4*a(n+1)*a(n+2) + 4*a(n+1)^2 = 0. - Emanuele Munarini, Dec 05 2017
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MAPLE
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q:=4; 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
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MATHEMATICA
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RecurrenceTable[{a[0]==0, a[1]==1, a[n]==a[n-1]+a[n-2]*4^(n-2)}, a, {n, 30}] (* Vincenzo Librandi, Nov 08 2012 *)
F[n_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^(j^2), {j, 0, Floor[(n-1)/2]}];
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PROG
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(Magma) [0] cat[n le 2 select 1 else Self(n-1) + Self(n-2)*(4^(n-2)): n in [1..20]]; // Vincenzo Librandi, Nov 08 2012
(PARI) q=4; 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
(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)))
(GAP) q:=4;; 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 16 2019
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CROSSREFS
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q-Fibonacci numbers: A000045 (q=1), A015459 (q=2), A015460 (q=3), this sequence (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).
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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