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 A049862 Products of two Fibonacci numbers with distinct indices. 6
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 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 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 *) PROG (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()); 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 CROSSREFS Cf. A000045, A160009, A272949. Sequence in context: A022826 A053035 A160009 * A022829 A347645 A229172 Adjacent sequences:  A049859 A049860 A049861 * A049863 A049864 A049865 KEYWORD nonn AUTHOR EXTENSIONS Name changed to conform with A272949 et al. by Clark Kimberling, Jun 18 2016 STATUS approved

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Last modified January 17 14:51 EST 2022. Contains 350401 sequences. (Running on oeis4.)