A135952 Prime factors of composite Fibonacci numbers with prime indices (cf. A050937).

37, 73, 113, 149, 157, 193, 269, 277, 313, 353, 389, 397, 457, 557, 613, 673, 677, 733, 757, 877, 953, 977, 997, 1069, 1093, 1153, 1213, 1237, 1453, 1657, 1753, 1873, 1877, 1933, 1949, 1993, 2017, 2137, 2221, 2237, 2309, 2333, 2417, 2473, 2557, 2593, 2749, 2777, 2789, 2797, 2857, 2909, 2917, 3217, 3253, 3313, 3517, 3557, 3733, 4013, 4057, 4177, 4273, 4349, 4357, 4513, 4637, 4733, 4909, 4933

Offset: 1

Author

Artur Jasinski, Dec 08 2007

Keywords

• nonn

Comments

All numbers in this sequence are congruent to 1 mod 4. - Max Alekseyev.

If Fibonacci(n) is divisible by a prime p of the form 4k+3 then n is even. To prove this statement it is enough to show that (1+sqrt(5))/(1-sqrt(5)) is never a square modulo such p (which is a straightforward exercise).

The n-th prime p is an element of this sequence iff A001602(n) is prime and A051694(n)=A000045(A001602(n))>p. - Max Alekseyev

Links

• Hans Havermann, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000045, A001605, A050937, A075737, A090819, A134787, A134851, A134852.

Programs

Mathematica

a = {}; k = {}; Do[If[ !PrimeQ[Fibonacci[Prime[n]]], s = FactorInteger[Fibonacci[Prime[n]]]; c = Length[s]; Do[AppendTo[k, s[[m]][[1]]], {m, 1, c}]], {n, 2, 60}]; Union[k]

Extensions

Edited, corrected and extended by Max Alekseyev, Dec 12 2007