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A252503
Smallest prime p such that Phi_n(p) is also prime, where Phi is the cyclotomic polynomial, or 0 if no such p exists.
1
3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 5, 2, 2, 2, 2, 2, 2, 7, 2, 11, 3, 2, 113, 2, 43, 2, 2, 5, 151, 2, 2, 2, 2, 2, 179, 3, 61, 2, 23, 2, 53, 2, 89, 137, 11, 2, 5, 5, 2, 7, 73, 11, 307, 7, 7, 2, 5, 7, 19, 3, 2, 2, 3, 0, 2, 53, 491, 197, 2, 3, 3, 3, 11, 19, 59, 7, 2, 2, 271, 2, 191, 61, 41, 7, 2, 2, 59, 5, 2, 2
OFFSET
1,1
COMMENTS
Conjecture: a(n) > 0 for all n != 2^k (k>5).
Clearly, if n is a power of 2, and Phi_n(2) is not prime, then a(n) = 0.
Records: 3, 5, 7, 11, 113, 151, 179, 307, 491, 839, 1427, 2411, 5987, 6389, 8933, 11813, 18587, 31721, 40763, 46349, ..., . - Robert G. Wilson v, May 21 2017
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 1..1300 (first 1000 terms from Charles R Greathouse IV)
MATHEMATICA
Do[n=1; p=Prime[n]; cp=Cyclotomic[k, p]; While[!PrimeQ[cp], n=n+1; p=Prime[n]; cp=Cyclotomic[k, p]]; Print[p], {k, 1, 300}]
PROG
(PARI) a(n)=if(n>>valuation(n, 2)==1 && n>32, if(ispseudoprime(2^(n/2)+1), 2, 0), my(P=polcyclo(n)); forprime(p=2, , if(ispseudoprime(subst(P, 'x, p)), return(p)))) \\ Charles R Greathouse IV, Dec 18 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Eric Chen, Dec 18 2014
STATUS
approved