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
A279543
a(n) = a(n-1) + 3^n * a(n-2) with a(0) = 1 and a(1) = 1.
Original entry on oeis.org
1, 1, 10, 37, 847, 9838, 627301, 22143007, 4137864868, 439978671649, 244776761262181, 78185678507867584, 130162592460442600405, 124783388108159412726037, 622688428086038843429228482, 1791127919536971393223950620041
Offset: 0
1/1 = a(0)/A015460(2).
1/(1+3/1) = 1/4 = a(1)/A015460(3).
1/(1+3/(1+3^2/1)) = 10/13 = a(2)/A015460(4).
1/(1+3/(1+3^2/(1+3^3/1))) = 37/121 = a(3)/A015460(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:
A280294 (q=2), this sequence (q=3),
A280340 (q=10).
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RecurrenceTable[{a[n] == a[n - 1] + 3^n*a[n - 2], a[0] == 1, a[1] == 1}, a, {n, 15}] (* Michael De Vlieger, Dec 31 2016 *)
A165902
a(0)=0, a(1)=1, a(n) = a(n-1) + 3^(n-3)*a(n-2).
Original entry on oeis.org
0, 1, 1, 2, 5, 23, 158, 2021, 40415, 1513724, 89901329, 10021444493, 1779549303200, 593535825170357, 315835356239140757, 315745107820598835194, 503859317773076207957705, 1510702921925354866376968691, 7231341194731242966764145947126
Offset: 0
q-Fibonacci numbers:
A000045 (q=1),
A165901 (q=2), this sequence (q=3).
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q:=3;; 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 19 2019
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q:=3; I:=[0,1]; [n le 2 select I[n] else Self(n-1) + q^(n-3)*Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 19 2019
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q:=3; seq(add((product((1-q^(n-j-1-k))/(1-q^(k+1)), k=0..j-1))* q^(j*(j-1)), j = 0..floor((n-1)/2)), n = 0..20); # G. C. Greubel, Dec 19 2019
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RecurrenceTable[{a[0]==0,a[1]==1,a[n]==a[n-1]+3^(n-3) a[n-2]},a,{n,20}] (* Harvey P. Dale, Oct 18 2014 *)
F[n_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^(j*(j-1)), {j,0,Floor[(n-1)/2]}]; Table[F[n, 3], {n,0,20}] (* G. C. Greubel, Dec 19 2019 *)
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a(n)=if(n<2,n,a(n-1)+3^(n-3)*a(n-2));
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def F(n,q): return sum( q_binomial(n-j-1, j, q)*q^(j*(j-1)) for j in (0..floor((n-1)/2)))
[F(n,3) for n in (0..20)] # G. C. Greubel, Dec 19 2019
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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