A283364 -id:A283364 - cp's OEIS frontend

A066263 Numbers k such that 2^k + 1 has just two distinct prime factors.

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

Offset: 1

Benoit Cloitre, Dec 31 2001

nonn

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 Mihăilescu-Catalan's theorems (see links), one can verify that any nonsquarefree term greater 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 ((Fj-1)^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)

			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.

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.

Cf. A073936, A107036, A000978, A057182, A127317, A283657, A283364, A092559, A046799.

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 *)

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

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

Edited by Robert G. Wilson v, Jan 03 2002

a(40)-a(52) by Giuseppe Coppoletta, May 02 2017

A283657 Numbers m such that 2^m + 1 has at most 2 distinct prime factors.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 16, 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

Offset: 1

Vladimir Shevelev, Mar 13 2017

nonn

Using comment in A283364, note that if a(n) is odd > 9, then it is prime.
503 <= a(41) <= 596. - Robert Israel, Mar 13 2017
Could (4^p + 1)/5^t be prime, where p is prime, 5^t is the highest power of 5 dividing 4^p + 1, other than for p=2, 3 and 5? - Vladimir Shevelev, Mar 14 2017
In his message to seqfans from Mar 15 2017, Jack Brennen beautifully proved that there are no more primes of such form. From his proof one can see also that there are no terms of the form 2*p > 10 in the sequence. - Vladimir Shevelev, Mar 15 2017
Where A046799(n)=2. - Robert G. Wilson v, Mar 15 2017
From Giuseppe Coppoletta, May 16 2017: (Start)
The only terms that are not in A066263 are those m giving 2^m + 1 = prime (i.e. m = 0 and any number m such that 2^m + 1 is a Fermat prime) and the values of m giving 2^m + 1 = power of a prime, giving m = 3 as the only possible case (by Mihăilescu-Catalan's result, see links).
For the relation with Fermat numbers and for other possible terms to check, see comments in A073936 and A066263.
All terms after a(59) 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(65), the values 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 further along in the numbering (see comments in A000978).
(End).

			0 is a term as 2^0 + 1 = 2 is a prime.
10 is a term as 2^10 + 1 = 5^2 * 41.
14 is not a term as 2^14 + 1 = 5 * 29 * 113.

Giuseppe Coppoletta, Table of n, a(n) for n = 1..65
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.

Cf. A002587, A046799, A283364, A073936, A000978, A127317, A092559, A019434, A066263.

Contains 4*A057182.

# this uses A002587[i] for i<=500, e.g., from the b-file for that sequence
count:= 0:
for i from 0 to 500 do
  m:= 0;
  r:= (2^i+1);
  if i::odd then
    m:= 1;
    r:= r/3^padic:-ordp(r,3);
  elif i > 2 then
    q:= max(numtheory:-factorset(i));
    if q > 2 then
      m:= 1;
      r:= r/B[i/q]^padic:-ordp(r,A002587[i/q]);
    fi
  fi;
  if r mod B[i] = 0 then m:= m+1;
      j:= padic:-ordp(r, A002587[i]);
      r:= r/B[i]^j;
  fi;
  mmax:= m;
  if isprime(r) then m:= m+1; mmax:= m
  elif r > 1 then mmax:= m+2
  fi;
  if mmax <= 2 or (m <= 1 and m + nops(numtheory:-factorset(r)) <= 2) then
       count:= count+1;
     A[count]:= i;
  fi
od:
seq(A[i],i=1..count); # Robert Israel, Mar 13 2017

Select[Range[0, 313], PrimeNu[2^# + 1]<3 &] (* Indranil Ghosh, Mar 13 2017 *)

for(n=0, 313, if(omega(2^n + 1)<3, print1(n,", "))) \\ Indranil Ghosh, Mar 13 2017

a(16)-a(38) from Peter J. C. Moses, Mar 13 2017
a(39)-a(40) from Robert Israel, Mar 13 2017
a(41)-a(65) from Giuseppe Coppoletta, May 08 2017

A283455 Numbers m such that 2^m - 1 has at most 2 distinct prime factors.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 17, 19, 23, 31, 37, 41, 49, 59, 61, 67, 83, 89, 97, 101, 103, 107, 109, 127, 131, 137, 139, 149, 167, 197, 199, 227, 241, 269, 271, 281, 293, 347, 373, 379, 421, 457, 487, 521, 523, 607, 727, 809, 881, 971, 983, 997, 1061

Offset: 1

Vladimir Shevelev, Mar 08 2017

nonn

The sequence differs from A283364 beginning with a(15). All a(n) > 6 are primes or squares of primes.
As in A283364 one can prove that all a(n) > 6 are odd. It is clear that a(n) is either prime or semiprime. Let us show that in the latter case it is the square of a prime. Indeed, let a(n) = p*q, p < q. Then 2^a(n)-1 is divisible by 2^p-1 < 2^q-1. Thus both of them are Mersenne primes.
Let us show that 2^(p*q)-1 differs from (2^p-1)^u*(2^q-1)^v, u,v >= 1. Indeed the equality is possible only in the case p*u + q*v = p*q. Then p|v and q|u. Let u = q*a, v = p*b. Then a + b = 1, which is impossible for u,v >= 1. Hence, 2^(p*q)-1 has a third prime divisor and p*q is not a member.
Are there terms other than 4, 9 and 49 that are squares of primes? Note that, for prime p, 2^(p^2)-1 differs from (2^p-1)^p, so if p^2 is a term, then for a Mersenne prime 2^p-1 and some t >= 1, the number (2^(p^2)-1)/(2^p-1)^t should be a prime or a power of a prime.
Numbers n such that A046800(n) < 3. - Michel Marcus, Mar 08 2017

The Cunningham Project, The Main Tables, Table 2- Factorizations of 2^n-1, n odd, n<1300.

Cf. A046800, A283364.

Union of {1}, A000043, A085724.

Select[Join[Range@ 6, Range[7, 201, 2]], PrimeNu[2^# - 1] <= 2 &] (* Michael De Vlieger, Mar 08 2017 *)

is(n)=omega(2^n-1)<3 \\ Charles R Greathouse IV, Mar 08 2017

More terms from Peter J. C. Moses, Mar 08 2017
a(48)-a(50) from Charles R Greathouse IV, Mar 08 2017
a(51)-a(57) from Amiram Eldar, Feb 13 2020

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A066263 Numbers k such that 2^k + 1 has just two distinct prime factors.

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A283657 Numbers m such that 2^m + 1 has at most 2 distinct prime factors.

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A283455 Numbers m such that 2^m - 1 has at most 2 distinct prime factors.

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