

A066263


Numbers k such that 2^k + 1 has just two distinct prime factors.


5



5, 6, 7, 9, 10, 11, 12, 13, 17, 19, 20, 23, 28, 31, 32, 40, 43, 61, 64, 79, 92, 101, 104, 127, 128, 148, 167, 191, 199, 256, 313, 347, 356, 596, 692, 701, 1004, 1228, 1268, 1709, 2617, 3539, 3824, 5807, 10501, 10691, 11279, 12391, 14479, 42737, 83339, 95369
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

From Giuseppe Coppoletta, May 16 2017: (Start)
All terms after a(52) refer to probabilistic primality tests for 2^a(n) + 1 (see Caldwell's link for the list of the largest certified Wagstaff primes). After a(58), 267017, 269987, 374321, 986191, 4031399 and 4101572 are also terms, but there still remains the remote possibility of some gaps in between. In addition, 13347311 and 13372531 are also terms, but possibly much farther in the numbering (see comments in A000978).
For the relation with Fermat numbers and for the primality of odd terms, see comments in A073936. The terms 9 and 10 give a value of 2^n + 1 which is not squarefree, so they are not in A073936. For the rest, the actually known terms of the two sequences coincide. In order to verify if any other term could be found hereafter that is not in A073936, all we have to do is to examine the terms for which 2^n + 1 is not squarefree. Considering that 3 divides 2^a(n) + 1 for any odd term a(n) and using Zsigmondy's and MihăilescuCatalan's theorems (see links), one can verify that any nonsquarefree term greather than 9 has to be of the form a(n) = 2^j * Fj, where Fj is the Fermat prime 2^2^j + 1. So basically we have to see if ((Fj1)^Fj + 1)/(Fj)^2 is a prime or the power of a prime for any Fermat prime Fj. The case j = 1 gives the term a(n) = 10 because (4^5 + 1)/5^2 = 41 is a prime, while for j = 2, (16^17 + 1)/17^2 = 354689 * 2879347902817 is composite. Similarly (256^257 + 1)/257^2 is neither a prime nor the power of a prime, so there is no contribution from the cases j = 2, 3 (see also comments in A127317).
For j = 4 and for any possible other Fermat prime which could be found later, the question is still open, in the sense that it is not actually known if n = 16 * F4 = 1048592 is a term or not. That seems very unlikely, but in order to decide that question for j = 4, one would have to check if (2^1048592 + 1)/65537^2 is a prime or the power of a prime. As this number has 315649 digits, I wonder if it is possible to handle it with the existing primality tests.
(End)


LINKS

Giuseppe Coppoletta, Table of n, a(n) for n = 1..63
Jack Brennen, Primes of the form (4^p+1)/5^t, Seqfan (Mar 15 2017).
C. Caldwell's The Top Twenty Wagstaff primes.
Mersennewiki, Factorizations Of Cunningham Numbers C+(2,n) (tables).
Samuel S. Wagstaff, The Cunningham Project.
Eric Weisstein's World of Mathematics, Catalan's Conjecture.
Eric Weisstein's World of Mathematics, Zsigmondy Theorem.


FORMULA

A001221(2^a(n) + 1) = 2.


EXAMPLE

3 and 4 are not terms because 2^3 + 1 and 2^4 + 1 have only a single prime factor (counted without multiplicity).
6 and 10 are terms because 2^6 + 1 = 5 * 13 and 2^10 + 1 = 5^2 * 41 have two distinct prime factors.


MATHEMATICA

f[n_] := First[ Transpose[ FactorInteger[2^n + 1]]]; Select[ Range[100], Length[f[ # ]] == 2 & ]
Select[Range[1300], PrimeNu[2^#+1]==2&] (* Harvey P. Dale, Nov 28 2014 *)


PROG

(PARI) isok(k) = #factor(2^k+1)~ == 2; \\ Michel Marcus, Nov 14 2017


CROSSREFS

Cf. A073936, A107036, A000978, A057182, A127317, A283657, A283364, A092559, A046799.
Sequence in context: A205857 A196026 A191850 * A178096 A246782 A089192
Adjacent sequences: A066260 A066261 A066262 * A066264 A066265 A066266


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Dec 31 2001


EXTENSIONS

Edited by Robert G. Wilson v, Jan 03 2002
a(40)a(52) by Giuseppe Coppoletta, May 02 2017


STATUS

approved



