

A250203


Numbers n such that the Phi_n(2) is the product of exactly two primes and is divisible by 2n+1.


0



11, 20, 23, 35, 39, 48, 83, 96, 131, 231, 303, 375, 384, 519, 771, 848, 1400, 1983, 2280, 2640, 2715, 3359, 6144, 7736, 7911, 11079, 13224, 16664, 24263, 36168, 130439, 406583
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OFFSET

1,1


COMMENTS

Here Phi_n is the nth cyclotomic polynomial.
Is this sequence infinite?
Phi_n(2)/(2n+1) is only a probable prime for n > 16664.
a(33) > 2000000.
Subsequence of A005097 (2 * a(n) + 1 are all primes)
Subsequence of A081858.
2 * a(n) + 1 are in A115591.
Primes in this sequence are listed in A239638.
A085021(a(n)) = 2.
All a(n) are congruent to 0 or 3 (mod 4). (A014601)
All a(n) are congruent to 0 or 2 (mod 3). (A007494)
Except the term 20, all even numbers in this sequence are divisible by 8.


LINKS

Table of n, a(n) for n=1..32.
Eric Chen, Gord Palameta, Factorization of Phi_n(2) for n up to 1280
Will Edgington, Factorization of completely factored Phi_n(2) [from Internet Archive Wayback Machine]
Henri Lifchitz and Renaud Lifchitz, PRP records. Search for (2^a1)/b
Samuel Wagstaff, The Cunningham project


EXAMPLE

Phi_11(2) = 23 * 89 and 23 = 2 * 11 + 1, so 11 is in this sequence.
Phi_35(2) = 71 * 122921 and 71 = 2 * 35 + 1, so 35 is in this sequence.
Phi_48(2) = 97 * 673 and 97 = 2 * 48 + 1, so 48 is in this sequence.


MATHEMATICA

Select[Range[10000], PrimeQ[2*# + 1] && PowerMod[2, #, 2*# + 1] == 1 &&
PrimeQ[Cyclotomic[#, 2]/(2*#+1)] &]


PROG

(PARI) isok(n) = if (((x=polcyclo(n, 2)) % (2*n+1) == 0) && (omega(x) == 2), print1(n, ", ")); \\ Michel Marcus, Mar 13 2015


CROSSREFS

Cf. A239638, A085724, A072226, A005384, A005097, A081858, A014601, A014664, A001917, A115591.
Sequence in context: A279431 A230541 A053715 * A339307 A038581 A044054
Adjacent sequences: A250200 A250201 A250202 * A250204 A250205 A250206


KEYWORD

nonn,more,hard


AUTHOR

Eric Chen, Mar 13 2015


STATUS

approved



