%I #48 Apr 30 2021 23:09:24

%S 0,1,3,20,52,357,935,6408,16776,114985,301035,2063324,5401852,

%T 37024845,96932303,664383888,1739379600,11921885137,31211900499,

%U 213929548580,560074829380,3838809989301,10050135028343,68884650258840,180342355680792,1236084894669817

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

%C 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.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,17,0,17,0,-1).

%F 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.

%F 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

%e 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.

%e 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.

%e 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.

%p F:= n-> (<<0|1>, <1|1>>^n)[1, 2]:

%p a:= n-> (k-> mul(F(k+j), j=0..2)/add(F(k+j), j=0..2))(floor(3*n/2)-1):

%p seq(a(n), n=1..30); # _Alois P. Heinz_, Apr 02 2021

%t Select[Table[(Fibonacci[k-1]*Fibonacci[k]*Fibonacci[k+1])/(Fibonacci[k-1]+Fibonacci[k]+Fibonacci[k+1]),{k,37}],IntegerQ] (* or *)

%t 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 *)

%o (PARI)

%o r(m)={fibonacci(m)*fibonacci(m-1)*fibonacci(m+1)/(fibonacci(m)+fibonacci(m-1)+fibonacci(m+1))}

%o { for(m=2, 30, my(t=r(m)); if(!frac(t), print1(t, ", ")))} \\ _Andrew Howroyd_, Apr 02 2021

%Y Cf. A000045 (Fibonacci numbers), 1/2 times the even terms of sequence A001654.

%Y Cf. A065563 (F(n-1)*F(n)*F(n+1)), A078642 (F(n-1)+F(n)+F(n+1)).

%K nonn,easy

%O 1,3

%A _Amrit Awasthi_, Apr 02 2021