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

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

Author

Amrit Awasthi, Apr 02 2021

Keywords

• nonn
• easy

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

• Index entries for linear recurrences with constant coefficients, signature (0,17,0,17,0,-1).

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