cp's OEIS Frontend

All terms are primes. Corresponding primes of the form (2^n + 11^n)/13 are {103, 12391, 38880540653801911, ...}.

a(11) > 10^5. - Robert Price, Feb 26 2013

The first 5 terms are primes.

Intersection of A000043 and A000978.

'The New Mersenne Conjecture' (Bateman-Selfridge-Wagstaff) states that if two of the following statements about an odd positive integer p are true, then the third one is also true: (a) p = 2^k +- 1 or p = 4^k +- 3, (b) 2^p - 1 is prime, (c) (2^p + 1)/3 is prime. (Amer Math Monthly, 96 (1989) p. 125.) - R. K. Guy, May 20 2005

The next term, if it exists, is not any currently known Mersenne prime exponent or Wagstaff prime exponent: it must be larger than A000043(48) = 57885161 and cannot be 74207281, 77232917, or 82589933. See Caldwell and both Wanless links. The New Mersenne Conjecture would require this sequence to be a subsequence of A122834, in which case the next term could not be less than A122834(28) = 2305843009213693951. See Caldwell and Höglund links. - Gord Palameta, Jun 28 2019, Jun 29 2024

p either has the form 2^k -+ 1 or the form 4^k -+ 3, according to the New Mersenne Conjecture. - Lekraj Beedassy, Sep 20 2006

Primes p such that (4^p - 1)/3 is a semiprime. - Arkadiusz Wesolowski, Jun 01 2013

Numbers m != 4 such that (4^m - 1)/3 is a semiprime. - Thomas Ordowski, Sep 25 2015

The indices of Wagstaff primes relating to the new Mersenne conjecture A122834 in a list of Jacobsthal numbers A001045. - Steve Homewood, Dec 01 2020

Let p be a prime in this sequence. Call q=2^p-1 and r=(2^p+1)/3. The new Mersenne conjecture implies that either q and r are both prime or both composite.

Does 2 occur infinitely often in A124151?

The sum in A124151 is 1+n if k=1, and 1+k*(k^(2n)-1)/(k^2-1) if k>1. The indices of A124151(n)=2 are where k=1 is avoided, but where k=2 leads to a prime, i.e., where 1+n is not prime but 1+2*(4^n-1)/3 = (2^(2n+1)+1)/3 is prime. Therefore this sequence here is constructed by taking all n=(A000978(i)-1)/2 (the members of A127936), and eliminating cases with 1+n in A000040. - R. J. Mathar, Feb 03 2010

All terms are primes. Largest currently known prime of the form (2^n + 1)/257 is (256^23029 + 1)/257 found by Donovan Johnson 03/2005. The only currently known prime of the form (2^n + 1)/65537 is (65536^239 + 1)/65537.

From Giuseppe Coppoletta, May 18 2017: (Start)

In general, for any j > 1, if (2^(n*2^j) + 1)/Fj is a prime (where Fj = 2^2^j + 1 is the corresponding Fermat number), then n needs to be prime, as for any odd proper factor q of n, 2^(q*2^j) + 1 is another factor of the numerator. The same for j = 0, apart for the particular value n = 3^2.

For the case j = 4, I checked it again, and (65536^p + 1)/65537 indeed is not a prime at least for 239 < p < 12500, i.e. (2^n + 1)/65537 is not a prime at least up to n = 200000. Any higher upper bound available?

One can also remark that 65536 = 2^16 and 239 = 2^8 - 2^4 - 1. Is there any special reason (see Brennen's link) for that?

I checked also that (2^(p*2^j) + 1)/Fj is never a proper power (in particular it is not a prime power) for j = 0..4 and for any prime p, at least for any exponent p*2^j < 200000.

We can even conjecture that ((Fj-1)^p + 1)/Fj is always squarefree for any odd prime p and for any Fermat number Fj with j >= 0. Note that this is not true if p is not restricted to be a prime, even if p and Fj are coprime, as shown by the following counterexample relative to the case j = 1, f1 = 5: 4^91 + 1 == 0 mod 1093^2. Remark that any such counterexample has to be a Wieferich prime (A001220), but not every Wieferich prime gives a counterexample, as shown by the second known Wieferich prime (3511), which cannot match here because it belongs to A072936.

(End)

These (2^k+1)/3 numbers have no prime factors of the form 2 (mod 3) to an odd power.

All terms are prime.

a(7) > 10^5. - Robert Price, Aug 26 2013

All terms are primes.

a(11) > 10^5.

All terms are odd primes.

a(38),...,a(43) = {3,3,7,13,7,7}.

a(46),...,a(64) = {19,3,3,13,13,11,19,41,3,7,109,11,227,13,5,5,3,239,5}.

a(66) = 3. a(68),...,a(72) = {3,7,31,3,7}.

a(74),...,a(92) = {3,5,19,17,3,83,3,3,19,19,11,11,61,3,7,7,3,41,29}.

a(94) = 5. a(97),a(98) = {19,7}. a(100) = 31.

a(n) is currently unknown for n = {37,44,45,65,67,73,93,95,96,99,...}.

From Kevin P. Thompson, May 18 2022: (Start)

All known terms from n=1..100 correspond to proven primes.

a(96) > 10250.

a(99) > 10250. (End)

Presuming every prime is seen at least once, one can specifically seek those with fixed k. Doing this, a(174) = 37, a(368) = 43 for example. - Bill McEachen, Aug 26 2024