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A085726
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Numbers n such that n-th Lucas number is a semiprime.
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4
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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MATHEMATICA
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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 *)
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PROG
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(Magma) IsSemiprime:=func<n | &+[k[2]: k in Factorization(n)] eq 2>; [n: n in [2..300] | IsSemiprime(Lucas(n))]; // Vincenzo Librandi, Feb 12 2016
(PARI) isok(n) = bigomega(fibonacci(n+1)+fibonacci(n-1)) == 2; \\ Michel Marcus, Feb 12 2016
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CROSSREFS
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Cf. A072381 (n such that Fibonacci(n) is a semiprime).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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More terms from Mark Hudson (mrmarkhudson(AT)hotmail.com), Aug 25 2004
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STATUS
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approved
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