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Numbers k such that F(3*k)/(2*F(k)) is prime, where F(m) is the m-th Fibonacci number.
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%I #31 Nov 06 2024 18:07:29

%S 5,7,11,13,17,31,37,41,67,107,151,257,349,457,787,911,1289,1627,3271,

%T 8233,13163,14551,31517,55579,103289

%N Numbers k such that F(3*k)/(2*F(k)) is prime, where F(m) is the m-th Fibonacci number.

%C All terms are primes. Conjecture: this sequence is infinite.

%e For n = 5 we have F(3*5)/(2*F(5)) = F(15)/(2*5) = 610/10 = 61 is prime.

%t Select[Range[1000], PrimeQ[Fibonacci[3*#]/Fibonacci[#]/2] &] (* _Vaclav Kotesovec_, Jul 18 2013 *)

%o (PARI) forprime(p=5,1e4,if(ispseudoprime(t=fibonacci(3*p)/fibonacci(p) /2), print1(p", "))) \\ _Charles R Greathouse IV_, Jul 16 2013

%o (PFGW) ABC2 F(3*$a)/2/F($a)

%o a: primes from 5 to 25000

%o // _Charles R Greathouse IV_, Jul 16 2013

%Y Cf. A001605, A000045, A227627.

%K nonn,more,changed

%O 1,1

%A _Thomas Ordowski_, Jul 16 2013

%E a(6)-a(22) from _Charles R Greathouse IV_, Jul 16 2013

%E a(23) from _Vaclav Kotesovec_, Jul 18 2013

%E a(24) from _Charles R Greathouse IV_, Jul 18 2013

%E a(25) from _Michael S. Branicky_, Nov 06 2024