1,1

Is it true that this sequence consists of the odd primes p with 2^2^p == 1 (mod p)? (David Wilson, Jul 31 2008). From Max Alekseyev: Yes! If prime p divides Fm = 2^(2^m)+1 then 2^(2^(m+1)) == 1 (mod p) and p is of the form p = k*2^(m+2)+1 > 2^(m+1). Taking squares p - 2^(m+1) times, we have 2^(2^p) == 1 (mod p). On the other hand, if 2^(2^p) == 1 (mod p) for prime p, consider a sequence 2^(2^0), 2^(2^1), 2^(2^2), ..., 2^(2^p). Modulo p this sequence ends with a bunch of 1's but just before the first 1 we must see -1 (as the only other square root of 1 modulo prime p), i.e. for some m, 2^(2^m) == -1 (mod p), implying that p divides Fermat number 2^(2^m) + 1.

Also primes p such that the multiplicative order of 2 (mod p) is a power of 2. A theorem of Lucas states that if m>1 and prime p divides 1+2^2^m (the m-th Fermat number), then p = 1+k*2^(m+2) for some integer k. [From T. D. Noe (noe(AT)sspectra.com), Jan 29 2009]

Wilfrid Keller analyzed the current status of the search for prime factors of Fermat number and stated that all prime factors less than 2^58 are now known. He sent me terms a(25) to a(44). [From T. D. Noe (noe(AT)sspectra.com), Feb 01 2009, Feb 03 2009]

M. Krizek, F. Luca & L. Somer, 17 Lectures on Fermat Numbers, Springer-Verlag NY 2001. [From T. D. Noe (noe(AT)sspectra.com), Jan 29 2009]

T. D. Noe, Table of n, a(n) for n=1..44 (from Wilfrid Keller)

Wilfrid Keller, Prime factors k.2^n + 1 of Fermat numbers F_m

R. Munafo, Prime Factors of Fermat Numbers

a(n) is a prime factor of the Fermat number 1+2^2^A023395(n). [From T. D. Noe (noe(AT)sspectra.com), Feb 01 2009]

Select[Prime[Range[100000]], IntegerQ[Log[2, MultiplicativeOrder[2, # ]]]&] [From T. D. Noe (noe(AT)sspectra.com), Jan 29 2009]

Cf. A000215.

Sequence in context: A083213 A056826 A058910 this_sequence A056130 A078726 A019434

Adjacent sequences: A023391 A023392 A023393 this_sequence A023395 A023396 A023397

nonn

David W. Wilson (davidwwilson(AT)comcast.net)