A015470
q-Fibonacci numbers for q=12, scaling a(n-2).
Original entry on oeis.org
0, 1, 1, 13, 157, 22621, 3278173, 5632106845, 9794204234077, 201818365309759837, 4211530365904119214429, 1041342647528423104910537053, 260767900948768868884822059725149, 773726564635922870118341112574642827613
Offset: 0
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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
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[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
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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
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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 *)
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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
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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
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
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).
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).
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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 *)
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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
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
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;
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.
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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
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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
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@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
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