A182504 Composite numbers k that divide both Fibonacci(k+1) and Fibonacci(2k+1)-1.

323, 377, 3827, 5777, 6479, 10877, 11663, 18407, 19043, 20999, 23407, 25877, 27323, 34943, 35207, 39203, 44099, 47519, 50183, 51983, 53663, 60377, 65471, 75077, 78089, 79547, 80189, 81719, 82983, 84279, 84419, 86063, 90287, 94667, 100127, 104663, 109871

Offset: 1

Author

Gary Detlefs, May 05 2012

Keywords

• nonn

Comments

A subset of A182554 based on a refinement of the Fibonacci criterion for primality described there. The additional constraint that k divides Fibonacci(2*k+1)-1 is suggested by the Cloitre comment in A003631.

What base-2 pseudoprimes are contained in this sequence?

An almost identical sequence can be obtained by testing for composite numbers for which (1) k divides Fibonacci(k+1) and (2) k^12 mod 210 = 1. All primes greater than 7 appear to satisfy condition 2. Terms of {a(n)} which are not pseudoprimes to this criterion are 50183, 65471, 82983, and 84279. - Gary Detlefs, Jun 04 2012

Links

• Giovanni Resta, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A182554, A003631, A038872, A000040.

Programs

Magma

[n: n in [4..11*10^4] | not IsPrime(n) and IsDivisibleBy(Fibonacci(n+1),n) and IsDivisibleBy(Fibonacci(2*n+1)-1,n)]; // Bruno Berselli, May 04 2012

Maple

with (combinat): f:= n-> fibonacci(n): for n from 2 to 100000 do if not isprime(n) and irem(f(n+1), n)=0 and irem((f(2*n+1)-1), n)=0 then print(n) fi od;

Mathematica

Select[Range[110000],CompositeQ[#]&&Mod[Fibonacci[#+1],#]==Mod[Fibonacci[ 2#+1]-1,#] == 0&] (* Harvey P. Dale, Aug 02 2024 *)