A085726 Numbers n such that n-th Lucas number is a semiprime.

3, 10, 14, 20, 23, 26, 29, 32, 38, 43, 49, 56, 62, 64, 67, 68, 73, 76, 80, 83, 86, 89, 97, 107, 121, 128, 136, 137, 157, 164, 167, 172, 178, 197, 202, 211, 223, 229, 284, 293, 307, 311, 328, 373, 389, 397, 458, 487, 521, 541, 557, 577, 586, 619, 673, 857, 914, 929, 947, 1082, 1151, 1249, 1277, 1279, 1306, 1318, 1493, 1499, 1667

Offset: 1

Author

Jason Earls, Jul 20 2003

Keywords

• nonn

Comments

From results on the divisibility of generalized Fibonacci sequences (2nd order recurrences with various integer initial values), it follows that if n is such that n-th Lucas number is a semiprime, it is necessary but not sufficient that n have at most two distinct prime factors (A070915). That is: A000204(n) an element of A001358 implies n an element of A070915. - Jonathan Vos Post, Sep 22 2005

All numbers in this sequence have the form 2^r p^s, where p is an odd prime and r and s are not both zero. It appears that s=2 for only p=7 and 11, otherwise s=0 or 1. - T. D. Noe, Nov 29 2005

Sequence continues as 1831?, 1877?, 1901, 1951, ... where ? mark uncertain terms. - Max Alekseyev, Aug 18 2013

Links

• Blair Kelly, Fibonacci and Lucas Factorizations

Crossrefs

Cf. A000204.

Cf. A072381 (n such that Fibonacci(n) is a semiprime).

Programs

Magma

IsSemiprime:=func; [n: n in [2..300] | IsSemiprime(Lucas(n))]; // Vincenzo Librandi, Feb 12 2016

Mathematica

a = 1; b = 3; Do[c = a + b; If[Plus@@Last/@FactorInteger[c] == 2, Print[n]]; a = b; b = c, {n, 3, 200}] (* Ryan Propper, Jun 28 2005 *)
Select[Range[400], PrimeOmega[LucasL[#]] == 2 &] (* Vincenzo Librandi, Feb 12 2016 *)

PARI

isok(n) = bigomega(fibonacci(n+1)+fibonacci(n-1)) == 2; \\ Michel Marcus, Feb 12 2016

Extensions

More terms from Mark Hudson (mrmarkhudson(AT)hotmail.com), Aug 25 2004

More terms from Ryan Propper, Jun 28 2005

More terms from T. D. Noe, Nov 29 2005

a(60)-a(62) from Max Alekseyev, Aug 18 2013

a(63)-a(69) from Sean A. Irvine, Feb 11 2016