A250203 Numbers n such that the Phi_n(2) is the product of exactly two primes and is divisible by 2n+1.

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

Offset

1,1

Comments

Here Phi_n is the n-th 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^a-1)/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: A230541 A053715 A343770 * A339307 A038581 A044054

Adjacent sequences: A250200 A250201 A250202 * A250204 A250205 A250206

Keyword

nonn,more,hard

Author

Eric Chen, Mar 13 2015

Status

approved