The Mersenne primes as the only sequence records of LPF(2^n - 1)

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Tomasz Ordowski

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Sep 3, 2026, 2:21:31 PMSep 3
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Hello Everyone, I have a trendy topic! 

Let us define the following sequence recursively:
a(1) = 2; a(n+1) is the smallest k such that 
lpf(2^k - 1) > lpf(2^a(n) - 1) for n > 0. 

2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, ? 

Are these only those primes p for which 2^p -1 is prime?

If so, then the Mersenne primes are the only sequence records 
of lpf(2^n - 1), and therefore also lpf(2^p - 1).  Read below...  


A049479. Smallest prime dividing 2^n - 1. A(n) = lpf(2^n - 1).
My new comment: 
It seems that the sequence records are only the Mersenne primes (for n in A000043).
See the sequence records of A016047 (for m in A016027). They are the same.
- Thomas Ordowski, Sep 01 2026 

Can this be accepted as presumed truth?
Find heuristic arguments for or against. 
Can this be called a conjecture? 

Best, 
Tom Ordo 
________________
I have some doubts.
By my recursive definition, the sequence (a(n)) is infinite and increasing, so it must have infinitely many records, but there need not be infinitely many Mersenne primes, although heuristics suggest so. 
Furthermore, it may turn out that the intervals between successive Mersenne exponents are too large for my guesses to hold true above p = 1279. 
What do you think? 

DONG HAOXUAN

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Sep 5, 2026, 12:33:38 AMSep 5
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Hi Tomasz,

I spent some time looking further into your conjecture and found a few useful reductions.

The main points are that composite indices cannot be strict records, every Mersenne-prime exponent is automatically a strict record, and the full conjecture would actually imply that there are infinitely many Mersenne primes.

I also looked more closely at the first difficult case, p = 1277. There are some interesting restrictions there, although I do not have a complete proof.

I have put the main arguments and results into the attached HTML file, since it is much easier to read there than in a long email.

I thought you might find it useful.

Best,
Jason Dong

main.html

Tomasz Ordowski

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Sep 5, 2026, 6:43:46 AMSep 5
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Dear Jason!

Thanks for taking the time to create your HTML file, which I read and understood easily. It will definitely come in handy again.

Firstly, it is very strange that no one has noticed this in the four centuries since Mersenne's work (1644), as far as I know. Perhaps it seemed like an obvious fact rather than a conjecture. 

Secondly, I wonder if this might have some loose connection to the New Mersenne Conjecture (1989), but I doubt it. 
What do you think?

Third and last, I will remind everyone of my conjecture: the records of lpf(2^n - 1) values are precisely the Mersenne primes, and their positions are the Mersenne exponents. Not to be confused with the positions of the records gpf(2^n - 1), A336721 (this is a "richer" sequence because it contains, in addition to all the Mersenne exponents, many other primes, together with composites 49 and 697). 
See https://oeis.org/A336721 (maybe someone will add a better comment; I am full, no free editing). 

Thanks for your attention. 

All the best!
Thomas Ordowski (3/3): 

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Antti Karttunen

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Sep 5, 2026, 7:35:41 AMSep 5
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I wonder whether Mersenne primes give also the only solutions mentioned in the conjecture in https://oeis.org/A364297 ?
And whether that has much to bear on the original topic here... Or maybe it's easy to solve.


Best regards,

Antti


M. F. Hasler

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Sep 5, 2026, 5:28:06 PMSep 5
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On Sat. 5 Sept. 2026, DONG HAOXUAN wrote:

Hi Tomasz, 

(...) I have put the main arguments and results into the attached HTML file, since it is much easier to read there than in a long email.


Nice writeup! Just a detail, you write, in your main.html sect.0.7

...proving

or, more weakly but sufficiently,

But I don't think this is weaker. If Omega(M) >= 3, the least of the three prime factors is at most M^(1/3),
(the maximum possible value is achieved if the three factors are ("nearly") equal, when one is larger then the least one is smaller), that is, lpf(M) <= M^(1/3). 
With M = Mp = 2^p-1,  p = 1277, this gives lpf(M) < 2^(p/3) < 2^427, which is much smaller than 2^607. 
So, for this M :   Omega(M) >= 3  implies  lpf(M) < M(607),  but not conversely,
which means that " Omega > 3" is the stronger and "lpf < ¨M(607)" is the weaker condition.

But maybe I got something wrong...

(Also, I always hesitate for a second to be sure whether lpf means least or largest prime factor. I think spf for smallest and gpf for greatest prime factor would be a better choice avoiding this ambiguity. (Also in view of LCS, LIS = largest / longest increasing/common subsequence, LCP = longest common prefix,  and other abbreviations where L means largest or longest.)

And yes, it's not yet proven that there are infinitely many Mersenne primes, 
so I didn't even attempt to prove Thomasz' conjecture which would imply that,
and must therefore be expected to be more difficult to prove.

I also quickly checked that it's not obvious to disprove: at first sight one might think it is not probable to have two consecutive Mersenne primes so far away from each other that a product of two or more primes between the two could still fit in between, but then one easily sees that it is sufficient that their exponents are at a ratio p_{n+1} / p_n > 2, which is the case already for n=12 (127 vs. 521, ratio 4.1), n=14 (where Jason Dong's potential counter example lives) and 20, but even more so for n=31 (216091 ~ 756839 / 3.5).

That illustrates that a counter-example is not so unlikely to be found.
To me that's much more likely than to find a proof of the "conjecture" that no such composite 2^k-1 < M_{n+1} with spf(2^k-1) > M_n exist.
(I might even conj^H^H^H take a (risky) bet that there is one below M_32 ~ 2^756839...)

- Maximilian

Tomasz Ordowski

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Sep 5, 2026, 5:43:16 PMSep 5
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Antti,

I can't answer your question because I haven't had time to delve into your sequence and understand the meaning of the Conjecture.

In any case, my conjecture isn't easy to disprove, let alone prove, because it implies that there are infinitely many Mersenne primes.
I won't pretend to have good heuristics either. Prof. Michael Filaseta only wrote to me that it's reasonable.

However, the earlier conjecture I promoted here had a disadvantageous heuristic, and Filaseta expected counterexamples for Mersenne exponents p >= 1279 in the case of k = 8.
But today, Amiram Eldar found a nice composite counterexample n = 221 = 13*17 with k = 8, which I overlooked. Is this the only exception? I don't think so!

Best,
Thomas  

DONG HAOXUAN

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Sep 5, 2026, 10:37:23 PMSep 5
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Hi everyone,

Thank you very much for the comments, and for taking the time to read my write-up. I really appreciate the positive feedback.

Maximilian, you are right about the point in Section 0.7. I had the stronger/weaker relation backwards. I have corrected it in the revised version. The correction also makes the situation at p = 1277 a little clearer: if it were a counterexample, M_1277 would have to be semiprime (counting prime factors with multiplicity).

I also agree with your point that the large gaps between some consecutive Mersenne exponents mean that size alone does not rule out a counterexample. So p = 1277 really is a nontrivial case.

Tomasz, regarding the New Mersenne Conjecture, I do not currently see a direct structural connection. Both involve Mersenne-prime exponents, but the conditions seem to come from rather different mechanisms.

Antti, I had a similar impression about A364297. It is interesting that Mersenne primes appear there as well, but I do not yet see a direct implication between that conjecture and this record problem.

I have gone through the HTML again, corrected the point Maximilian noticed, and made a few other small clarifications. The main elementary results are unchanged.

Thanks again for the careful reading and helpful comments.

Best,
Jason Dong


main2.html

Tomasz Ordowski

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Sep 6, 2026, 12:43:47 AMSep 6
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Thanks to Maksymilian for his polemical voice of reason.
   A conjecture without good heuristics is just a poor guess.
We can end this thread here, unless Jason defends himself.
   Thank you all for participating in this informative discussion.

Until next time! 
   Thomas 

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Tomasz Ordowski

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Sep 6, 2026, 12:48:03 AMSep 6
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PS. Maximilian, sorry, I polonized your name.

Tomasz Ordowski

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Sep 7, 2026, 3:04:49 PMSep 7
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For those interested,
I am attaching a note
from Michael Filaseta. 
OrdowskiEmail06Sept026.pdf

Tomasz Ordowski

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Sep 8, 2026, 1:18:46 AMSep 8
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For those confused (not only my fault), the following explanation.

Filaseta's note concerns my sequence defined by the recursion

a(1) = 2; and then, for n > 0, a(n+1) is the least k such that
           lpf(2^k - 1) > lpf(2^a(n) - 1),  
where lpf(m) is the least prime factor of m. 

And my "reasonable conjecture" [sic] states that 
the terms of this sequence coincide with the Mersenne exponents. 

I hope everything is clear now. 

That’s about it.  

Thomas Ordowski 

Tomasz Ordowski

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Sep 14, 2026, 1:32:29 AMSep 14
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News! 

The recent discovery of the factorization of the Mersenne number 2^1277-1, 
which I stumbled upon while using FactorDB, 
confirms my conjecture up to the Mersenne exponent p = 1279, and further up to p = 4423 
because of the gaps between the Mersenne primes prior to that (as Filaseta pointed out to me).

I therefore ventured to formulate a similar conjecture for Wagstaff numbers 
and I request an independent verification. 

One could also attempt to generalize such conjectures (regarding record values of the lpf function) 
to numbers of the form (a^n ± 1)/(a ± 1) and, more broadly, to (a^n ± b^n)/(a ± b) for sufficiently large n. 

This provides a broader perspective on previous conjectures (a=2, b=1)
before significant progress is made in the factorization of large numbers.

Good luck! 
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