cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A226958 a(n) = Fibonacci(n-2)*Fibonacci(n)*Fibonacci(n+2).

Original entry on oeis.org

2, 0, 10, 24, 130, 504, 2210, 9240, 39338, 166320, 705058, 2985840, 12649570, 53582256, 226981610, 961503816, 4073004770, 17253510120, 73087065922, 309601740360, 1311494081482, 5555577978720, 23533806138050, 99690802301664, 422297015715650, 1788878864564064, 7577812474943050
Offset: 1

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Author

Ron Knott, Jun 27 2013

Keywords

Examples

			a(3) = F(1)*F(3)*F(5) = 1*2*5 = 10.

Links

Crossrefs

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers).
Products of 3 Fibonaccis: A065563, A056570, A220362, A110224.

Programs

  • Mathematica
    Table[Fibonacci[n - 2] Fibonacci[n] Fibonacci[n + 2], {n, 1, 20}]
LinearRecurrence[{3,6,-3,-1},{2,0,10,24},30] (* Harvey P. Dale, Apr 10 2022 *)
Join[{2},#[[1]]#[[3]]#[[5]]&/@Partition[Fibonacci[Range[0,40]],5,1]] (* Harvey P. Dale, May 20 2025 *)
  • PARI
    a(n)=fibonacci(n-2)*fibonacci(n)*fibonacci(n+2); \\ Joerg Arndt, Jul 07 2013

Formula

a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4).
G.f.: 2*(1-3*x-x^2)/(1-3*x-6*x^2+3*x^3+x^4).
a(n) = Lucas(n-1)*Fibonacci(n+2) = Fibonacci(n-2)*Lucas(n+1).
a(n) = (1/5)*(Fibonacci(3*n)-8*(-1)^n*Fibonacci(n)). - Ehren Metcalfe, Mar 26 2016
For n >= 3, a(n) is the numerator of the continued fraction [1,..,1, 3 ,1,..,1, 3 ,1,..,1] with three runs of 1's each of length n-3 and each separated by a single 3. For example, a(5)=130 which is the numerator of the continued fraction [1,1, 3 ,1,1, 3 ,1,1]. - Greg Dresden, Jan 01 2022

Extensions

More terms from Joerg Arndt, Jul 07 2013

A279890 Expansion of x*(1 - x + 2*x^3 - x^4)/((1 - x)*(1 + x)*(1 - x + x^2)*(1 - x - x^2)).

Original entry on oeis.org

0, 1, 1, 2, 4, 7, 12, 19, 31, 50, 82, 133, 216, 349, 565, 914, 1480, 2395, 3876, 6271, 10147, 16418, 26566, 42985, 69552, 112537, 182089, 294626, 476716, 771343, 1248060, 2019403, 3267463, 5286866, 8554330, 13841197, 22395528, 36236725, 58632253, 94868978, 153501232, 248370211, 401871444, 650241655, 1052113099, 1702354754
Offset: 0

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Author

Ilya Gutkovskiy, Dec 22 2016

Keywords

Comments

The integer part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2).
The o.g.f. for the numerators of the fractional part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2) is 6*x/((1 + x - x^2)*(1 - 4*x - x^2)).
The o.g.f. for the denominators of the fractional part of the harmonic mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2) is (1 + 3*x - x^2)/((1 + x)*(1 - 3*x + x^2)).
Convolution of Fibonacci numbers and periodic sequence [1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, ...].

Examples

			a(1) = floor(3/(1/F(1)+1/F(2)+1/F(3))) = floor(3/(1/1+1/1+1/2)) = 1;
a(2) = floor(3/(1/F(2)+1/F(3)+1/F(4))) = floor(3/(1/1+1/2+1/3)) = 1;
a(3) = floor(3/(1/F(3)+1/F(4)+1/F(5))) = floor(3/(1/2+1/3+1/5)) = 2, etc.

Links

Crossrefs

Cf. A000045, A001622, A004695, A065563, A236428.
Cf. A062114 (the integer part of the harmonic mean of Fibonacci(n+1) and Fibonacci(n+2) for n>0).
Cf. A074331 (the integer part of the geometric mean of Fibonacci(n), Fibonacci(n+1) and Fibonacci(n+2)).

Programs

  • Mathematica
    LinearRecurrence[{2, 0, -2, 2, 0, -1}, {0, 1, 1, 2, 4, 7}, 46]
Table[Floor[3 Fibonacci[n] Fibonacci[n + 1] Fibonacci[n + 2]/(2 Fibonacci[n + 1] Fibonacci[n + 2] - (-1)^n)], {n, 0, 45}]
  • PARI
    concat(0, Vec((x*(1-x+2*x^3-x^4)/((1-x)*(1+x)*(1-x+x^2))) + O(x^40))) \\ Felix Fröhlich, Dec 22 2016

Formula

G.f.: x*(1 - x + 2*x^3 - x^4)/((1 - x)*(1 + x)*(1 - x + x^2)*(1 - x - x^2)).
a(n) = 2*a(n-1) - 2*a(n-3) + 2*a(n-4) - a(n-6).
a(n) = (9*sqrt(5)*(((1 + sqrt(5))/2)^n - ((1 - sqrt(5))/2)^n) + 5*((-1)^n + 2*cos(Pi*n/3) - 3))/30.
a(n) = floor(3*F(n)*F(n+1)*F(n+2)/(2*F(n+1)*F(n+2)-(-1)^n)), where F(n) is the n-th Fibonacci number (A000045).
a(n) = floor(3*A065563(n)/A236428(n+1)).
a(n) = 3*A000045(n)/2 + ((-1)^n + 2*cos(Pi*n/3) - 3)/6.
a(n) ~ 3*phi^n/(2*sqrt(5)), where phi is the golden ratio (A001622).
Lim_{n->infinity} a(n+1)/a(n) = phi.

A343010 Integers k for which there exist three consecutive Fibonacci numbers a, b, and c such that a*b*c = k*(a+b+c).

Original entry on oeis.org

0, 1, 3, 20, 52, 357, 935, 6408, 16776, 114985, 301035, 2063324, 5401852, 37024845, 96932303, 664383888, 1739379600, 11921885137, 31211900499, 213929548580, 560074829380, 3838809989301, 10050135028343, 68884650258840, 180342355680792, 1236084894669817
Offset: 1

Views

Author

Amrit Awasthi, Apr 02 2021

Keywords

Comments

F(n-1)*F(n)*F(n+1) = k(n)*(F(n-1)+F(n)+F(n+1)). This implies that k(n)=(F(n-1)*F(n))/2. Now k(n) will be an integer only when n is of the form 3*m or 3*m+1. Therefore we get k = (F(3*m+-1)*F(3*m))/2.

Examples

			0 is a term because F(0)*F(1)*F(2)/(F(0)+F(1)+F(2)) is 0*1*1/(0+1+1) = 0.
1 is a term because F(2)*F(3)*F(4)/(F(2)+F(3)+F(4)) is 1*2*3/(1+2+3) = 1.
3 is a term because F(3)*F(4)*F(5)/(F(3)+F(4)+F(5)) is 2*3*5/(2+3+5) = 3.

Links

Crossrefs

Cf. A000045 (Fibonacci numbers), 1/2 times the even terms of sequence A001654.
Cf. A065563 (F(n-1)*F(n)*F(n+1)), A078642 (F(n-1)+F(n)+F(n+1)).

Programs

  • Maple
    F:= n-> (<<0|1>, <1|1>>^n)[1, 2]:
a:= n-> (k-> mul(F(k+j), j=0..2)/add(F(k+j), j=0..2))(floor(3*n/2)-1):
seq(a(n), n=1..30);  # Alois P. Heinz, Apr 02 2021
  • Mathematica
    Select[Table[(Fibonacci[k-1]*Fibonacci[k]*Fibonacci[k+1])/(Fibonacci[k-1]+Fibonacci[k]+Fibonacci[k+1]),{k,37}],IntegerQ] (* or *)
b[k_]:=Fibonacci[3k-1]*Fibonacci[3k]/2; c[k_]:=Fibonacci[3k+1]*Fibonacci[3k]/2; Union[Table[b[k],{k,0,12}],Table[c[k],{k,0,12}]] (* Stefano Spezia, Apr 03 2021 *)
  • PARI
    r(m)={fibonacci(m)*fibonacci(m-1)*fibonacci(m+1)/(fibonacci(m)+fibonacci(m-1)+fibonacci(m+1))}
{ for(m=2, 30, my(t=r(m)); if(!frac(t), print1(t, ", ")))} \\ Andrew Howroyd, Apr 02 2021

Formula

Union of the two sequences b(k) and c(k) defined respectively as F(3*k-1)*F(3*k)/2 and F(3*k+1)*F(3*k)/2.
G.f.: x^2*(1 + 3*x + 3*x^2 + x^3)/(1 - 17*x^2 - 17*x^4 + x^6). - Stefano Spezia, Apr 03 2021
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