Carmichael-Mersenne primes

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Tomasz Ordowski

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Aug 12, 2026, 5:14:23 AMAug 12
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Hello! 

Today I have something for prime number lovers. 

First, a provocative question: 

Is lambda(2^n - 1) + 1 prime for almost all n ?

The numbers n for which this expression is prime 
can be called Carmichael-Mersenne exponents. 

How to prove that their set is infinite?

The second question is serious. 

Best, 

Tom Ordo 
________________

Drumețul Dacic

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Aug 12, 2026, 7:31:06 AMAug 12
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Interesting question!

Here is some data for the problem: values of n, for which lambda(2^n - 1)+1 is prime:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 31, 32, 34, 37, 38, 39, 40, 45, 48, 49, 50, 51, 52, 54, 56, 57, 61, 62, 65, 66, 68, 72, 77, 81, 83, 85, 87, 89, 107, 122, 127, 131, 133, 138, 179, 182, 219, 234, 235, 254, 310, 350, 376, 429, 456, 521, 534

Computed with the following Sidef program:

Num!USE_FACTORDB = true
for n in (1..534) {
    if (lambda(2**n - 1) + 1 -> is_prime) {
       print(n, ", ")
    }
}
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DONG HAOXUAN

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Aug 12, 2026, 8:02:42 AMAug 12
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Hi Tom,

I continued playing with this and found a fun one: n = 729 also works.

C(729) = lambda(2^729 - 1) + 1

is a 174-digit prime. I also verified its primality with a Pocklington
certificate, so this is not just a probable-prime result.

What surprised me more is that 729 is a very strong external case. If
P_n is the largest prime factor of 2^n - 1 and

D(n) = lambda(2^n - 1)/(P_n - 1),

then D(729) has 85 digits. So this is not coming from one dominant
Mersenne factor; the full LCM itself is doing something interesting.

For n = 3^k, the first six cases are now

P, P, P, P, C, P

for n = 3, 9, 27, 81, 243, 729.

I thought this one was worth mentioning!

Best,
Jason
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Tomasz Ordowski

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Aug 12, 2026, 4:00:55 PMAug 12
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Thanks for participating in the discussion so far. I really appreciate it. 

Let a(n) = lambda(2^n - 1) + 1. Finally, one more question to consider:
Are there only finitely many pairs m,n such that a(n) = a(m) for n > m?
Does this equality always apply to prime numbers? It's worth knowing. 

Until next time, 

Tom Ordo 

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CABbU%3D%2BUByoEjN-6teYyN4UCsSpjKqW00H1BTsQ2XienTawguNQ%40mail.gmail.com.

DONG HAOXUAN

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Aug 12, 2026, 7:21:57 PMAug 12
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Hi Tom,

A quick update from my side — I’ve kept working on the problem and
there has been quite a bit of progress, although I still don’t have a
proof of infinitude yet.

A few useful structural results have come out of it. In particular, I
found a finite “Carmichael envelope” criterion for controlling when
λ(2^n - 1) divides a given integer, which also gives complete finite
classifications for fixed small values of λ(2^n - 1)/n. I’ve also been
looking more closely at the prime-exponent case, where the semiprime
situation seems to connect very naturally with the base-2 Fermat
quotient.

I’m now trying to attack the main infinitude question more directly,
so I’ll send another update if that goes anywhere interesting.

By the way, are you thinking of eventually writing this up as a paper
or short note, or is your main plan for now to develop the OEIS entry
and the SeqFan discussion?

Best,
Jason

On Wed, 12 Aug 2026 22:00:42 +0200, Tomasz Ordowski
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Tomasz Ordowski

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Aug 13, 2026, 4:24:33 AMAug 13
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PS. Try also: lambda(2^n + 1) + 1. 
Does it give more prime numbers? 

DONG HAOXUAN

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Aug 13, 2026, 4:33:57 AMAug 13
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Yeah sure, I'm currently looking for a solution, but haven't made any significant progress yet. And thank you for your suggestion; I'll give it a try.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAF0qcNN1mYeqjGPQOhMtJXjkYFebyV2iJNtv3P9ZqxHkzhLghw%40mail.gmail.com.

Drumețul Dacic

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Aug 13, 2026, 10:14:54 AMAug 13
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Values of n such that lambda(2^n + 1)+1 is prime:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 27, 28, 29, 30, 31, 32, 33, 36, 38, 39, 40, 41, 43, 49, 53, 57, 60, 61, 69, 78, 79, 85, 87, 92, 97, 101, 104, 121, 127, 131, 138, 140, 143, 148, 150, 161, 167, 180, 183, 186, 187, 191, 199


On Thursday, August 13, 2026, DONG HAOXUAN <donghaoxuan...@gmail.com> wrote:
Yeah sure, I'm currently looking for a solution, but haven't made any significant progress yet. And thank you for your suggestion; I'll give it a try.

On Thu, Aug 13, 2026 at 6:24 PM Tomasz Ordowski <tomaszo...@gmail.com> wrote:
PS. Try also: lambda(2^n + 1) + 1. 
Does it give more prime numbers? 

śr., 12 sie 2026 o 22:00 Tomasz Ordowski <tomaszo...@gmail.com> napisał(a):
Thanks for participating in the discussion so far. I really appreciate it. 

Let a(n) = lambda(2^n - 1) + 1. Finally, one more question to consider:
Are there only finitely many pairs m,n such that a(n) = a(m) for n > m?
Does this equality always apply to prime numbers? It's worth knowing. 

Until next time, 

Tom Ordo 

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Tomasz Ordowski

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Aug 13, 2026, 11:56:36 AMAug 13
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Great, thanks!
At the end: lambda(4^n - 1) + 1. 
The end crowns the work. 
Now let's think. 

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DONG HAOXUAN

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Aug 13, 2026, 7:28:55 PM (14 days ago) Aug 13
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Hi Tom,

I tried following one of the directions you suggested, involving
λ(2^n+1)+1, and I also explored several other possible approaches to
the original Carmichael–Mersenne problem.

There is indeed a very natural connection here. If I write

L_n = λ(2^n−1), B_n = λ(2^n+1),

then

L_{2n} = lcm(L_n, B_n),

since 2^{2n}−1 = (2^n−1)(2^n+1). Likewise,

λ(4^n−1) = lcm(L_n, B_n).

So I think your 2^n+1 and 4^n−1 suggestions are genuinely connected to
the original problem, rather than being unrelated variants.

However, after trying this and several other routes fairly seriously,
I kept reaching essentially the same obstruction. Even when the
Carmichael exponent can be described very precisely, the final step
asks us to prove that a highly structured quantity of the form

lcm(⋯)+1

is prime infinitely often. I have not found a way around this
prime-value barrier.

I also tried internal lifting, doubling chains, primitive-divisor
arguments, and a few Euclid-type constructions. They give useful
structural information, but so far none of them removes this final
difficulty. My feeling is that some of the other natural routes may
eventually run into the same obstruction as well.

I thought it was worth mentioning this now, mainly so that we do not
unknowingly duplicate too much work. I am certainly not saying the
other directions are useless — the 2^n+1 side in particular seems
useful for understanding how the two Carmichael exponents interact —
but at least the most direct infinitude arguments appear to hit the
same wall.

Best,
Jason Dong

On Thu, 13 Aug 2026 17:56:24 +0200, Tomasz Ordowski
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Tomasz Ordowski

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Aug 14, 2026, 1:40:47 AM (14 days ago) Aug 14
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A heuristic for the abundance of primes.

OEIS A263027, a(n) = lambda(n) + 1, is already remarkably rich in primes. This makes our present problem, concerning the much more structured sequences lambda(2^n − 1) + 1 and lambda(2^n + 1) + 1, a problem of a higher order.

For N = 2^n ± 1, lambda(N) is the lcm of p − 1 over the prime divisors p of N. Since the orders of 2 modulo these primes are constrained by n, lambda(N) is far from a random integer. This may explain the unusually high prime density observed in lambda(2^n ± 1) + 1, with the + case apparently stronger than the − case.

For 4^n − 1 = (2^n − 1)(2^n + 1), however,

lambda(4^n − 1) = lcm(lambda(2^n − 1), lambda(2^n + 1)).

The lcm may eliminate part of the arithmetic redundancy present in the two separate Carmichael values, bringing lambda(4^n − 1) + 1 much closer to the generic prime-density scale.

Thus a natural heuristic is C+ > C− > C0, where C0 describes the baseline behavior of lambda(n) + 1.

But can this phenomenon be proved at all? In particular, can one prove that lambda(2^n − 1) + 1, or lambda(2^n + 1) + 1, is prime for infinitely many n?

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