[seqfan] Re: Mersenne numbers that are de Polignac numbers

[Tomasz Ordowski]

Professor Michael Filaseta became interested in my post.
This is what he wrote back to me:

You are looking at a particular value of a in Crocker's theorem, and he was
interested in something more general.  For what you want, you can get some
information based on the idea behind Crocker's proof.  The numbers

27, 55, 111, ...

are simply the numbers of the form

7*2^k-1, for k > 1.

Notice that when you double one of these and add 1, you get the next one.

So you want to know if

2^(7*2^k-1) - 1 - 2^n

is composite for n > 0 and so necessarily divisible by an odd prime.
Doubling the number will not change the odd prime factors, and this can be
written as

N = (2^(7*2^k) - 1) - (2^(n+1) + 1).

Crocker's point is that if we write n+1 = u*2^r, where u is odd and r is a
nonnegative integer, then 2^(2^r) + 1 is a factor of 2^(n+1) + 1.  As
2^(2^k) - 1 is a factor of 2^(7*2^k) - 1 and

2^(2^k) - 1 = (2^(2^(k-1)) + 1)*(2^(2^(k-2)) + 1)*...*(2^(2^0) + 1),

he concludes that 2^(2^r) + 1 divides N and is less than N provided r < k.

So we may suppose r >= k.  Since also n+1 < 7*2^k, we are left with the 6
possibilities n+1 = j*2^k where 0 < j < 7 for possible primality.  I was
able to check that these 6 possibilities are composite for k up to 14.  But
we are talking about large numbers now where computations are essentially
impossible (k > 14 implies N has over 69000 digits).   The size of log N is
larger than 2^k, so the "expected" number of prime values as k varies is at
most sum over k > 14 of 1/2^k = 0.000061....  So it is reasonable that
these continue by randomness to be composite.  Keep in mind though, this is
not unexpected.  You are looking at numbers that are doubly exponential, so
this is similar to asking if the Fermat numbers are all composite starting
with 2^(2^5)+1.  Basically, one can ask if any sequence that grows doubly
exponentially is always composite if the initial small cases, say where the
numbers are < 10^100, are composite.  As another example, is the number

floor( e^(pi^n) )

composite for all n > 20?  The fact that the numbers are huge and growing
very quickly just makes the problem very very difficult and the answer is
likely to be yes.  But you can imagine how easy it is to ask similar
questions about sequences that grow quickly.



Michael


*_______________________________________________*



*Michael Filaseta*, *Carolina Trustee Professor Emeritus*

Mathematics Department

University of South Carolina



https://people.math.sc.edu/filaseta/


pt., 19 sty 2024 o 13:50 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

> Dear number lovers,
>
> I have something interesting!
>
> Let's define (notice the obvious binary interpretation of this
> definition):
> Numbers m > 1 such that 2^m-1 - 2^n is composite for every natural n < m.
> 7, 15, 23, 27, 31, 37, 39, 43, 55, 58, 63, 71, 79, 82, 91, 95, 111, 123,
> 127, ...
>
> All integers m = 2^k-1 for k > 2 are in this sequence (by Crocker
> theorem).*
>
> The following property is more general.
> For many terms m, 2m+1 is also in this sequence (iteration).
> So maybe b(n) = (m+1)2^n-1 for n >= 0 is an infinite subsequence.
> For which m can this be true? For m = 7, 15, 31, .. this is already
> proven.*
> It seems that (27+1)2^n-1 is a subsequence: 27, 55, 111, 223, 447, ...
>
> How to prove that there are infinitely many primes in this sequence?
>
> Unfortunately, not all numbers n == -1 (mod 8) are in this sequence.
> However, it should be noted that many terms m == -1 (mod 8).
> See https://oeis.org/A004771 (a(n) = 8n+7 for n >= 0).
>
> Cf. https://oeis.org/A006285 (de Polignac numbers)
> and https://oeis.org/A000225 (Mersenne numbers).
>
> Best,
>
> Tom Ordo
> ______________
> (*) Crocker (1971) proved that,
> if m > 2 and a <> b, then 2^(2^m)-1 - 2^a - 2^b is not prime.
> If a = 2^m-1, then b < a, so for m > 2, 2^(2^m-1)-1 is a de Polignac
> number.
> Note that 2^(2^m-1)-1 - 2^n is divisible by some prime factor of
> 2^(2^m)-1.
>
> _____Post Scriptum__________
> Note that 127^7 and (2^127-1)^3 are de Polignac numbers.
> Difficult question: is (2^127-1)^4 a dual Sierpinski number?
>
>