A049862 Products of two Fibonacci numbers with distinct indices.

0, 1, 2, 3, 5, 6, 8, 10, 13, 15, 16, 21, 24, 26, 34, 39, 40, 42, 55, 63, 65, 68, 89, 102, 104, 105, 110, 144, 165, 168, 170, 178, 233, 267, 272, 273, 275, 288, 377, 432, 440, 442, 445, 466, 610, 699, 712, 714, 715, 720, 754, 987, 1131

Offset: 1

Author

Clark Kimberling

Keywords

• nonn

Comments

There are no duplicates except for the trivial cases 1*F(j)=1*F(j) and F(i)*F(j)=F(j)*F(i). - Robert Israel, May 11 2016

The number 1 is included because 1 = F(1)*F(2). - Clark Kimberling, Jun 19 2016

Links

• T. D. Noe, Table of n, a(n) for n=1..1000
• Mohammad K. Azarian, The Value of a Series of Reciprocal Fibonacci Numbers, Problem B-1133, Fibonacci Quarterly, Vol. 51, No. 3, August 2013, p. 275. Solution published in Vol. 52, No. 3, August 2014, pp. 277-278.
• MathOverflow, Distinctness of products of Fibonacci numbers

Crossrefs

Cf. A000045, A160009, A272949.

Programs

Maple

fib:= combinat:-fibonacci:
sort(convert(select(`<`,{0,seq(seq(fib(i)*fib(j),i=j+1..100),j=1..100)},fib(101)),list)); # Robert Israel, May 11 2016

Mathematica

Take[Union[Flatten[Table[Fibonacci[i]*Fibonacci[j], {i, 0, 100}, {j, i + 1, 100}]]], 100] (* Clark Kimberling, May 11 2016 *)

PARI

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
isok(n) = {if ((n==0) || (n==1), return (1)); fordiv(n, d, if (d^2 < n, if (isfib(d) && isfib(n/d), return (1)););); return(0);} \\ Michel Marcus, May 27 2019

PARI

lista(nn) = {my(out = List([0])); for (i=0, nn, for (j=i+1, nn, listput(out, fibonacci(i)*fibonacci(j)););); Vec(vecsort(select(x->(x < fibonacci(nn+1)), out), , 8));} \\ Michel Marcus, May 27 2019

Extensions

Name changed to conform with A272949 et al. by Clark Kimberling, Jun 18 2016