The question of divisibility

[Tomasz Ordowski]

Hello! 

As is well known, n! | Product_{k=2..n} (2^k - 2). 

Cf. https://oeis.org/A091669 (see comments there). 

Similarly, (2n+1)!! | Product_{k=1..n} (2^(2k) - 1). 

Is p(n)!! | Product_{k=2..n} (2^(p(k)-1) - 1) 

for every n > 1, where p(n) = prime(n)? 

 

Best, 

 

Tom Ordo 

[Emmanuel Vantieghem]

Op do 9 jul 2026 at 14:07 Tomasz Ordowski <tomaszo...@gmail.com> asked :

. 

Is  p(n)!! | Product_{k=2..n} (2^(p(k)-1) - 1) 

for every n > 1, where p(n) = prime(n) ? 

I found the first counterexample for  n = 31 :

   Prime[31]!! = 1647492604360283009858821457422710203523523237653188150644527258446635\
71025238239569133424206748199462890625 = a 

   Product[(2^(Prime[k] - 1) - 1), {k, 2, n}] = 9512332930388318659697458614535054961638477070953420247003082728966623\

4823304054888393260094520022162400439541280867335501328293777053085337\
2473961610842545175080929871707650420804515681839311409124826670141476\
7904243988687870585063037065203904033724043770971852832629102635991042\
6933393293500446645862718640414168117618366291782118067369347272763394\
3290929169931154475032738746026918608446245306323935067590454825822345\
0539210445419888645416099420428498141357525924967315105629579220536583\
217975616455078125 = b

   Mod[b,a] = 5990882197673756399486623481537128012812811773284320547798280939805948\

0372813905297866699711544799804687500

Sincere greetings.

Emmanuel Vantieghem

 

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[Emmanuel Vantieghem]

The congruence is further on true for n = 32,33,34,36, 37  and no other value below 1229.

Op vrijdag 10 juli 2026 om 14:26:10 UTC+2 schreef Emmanuel Vantieghem:

[Tomasz Ordowski]

Emmanuel, well, thanks! 

However, I believe that 

(2n+1)!! | Product_{k=1..n} (2^(2k) - 1)

is provable. Right? 

Tom  

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[Emmanuel Vantieghem]

Op vr 10 jul 2026 om 20:22 schreef Tomasz Ordowski <tomaszo...@gmail.com>:

...

However, I believe that 

(2n+1)!! | Product_{k=1..n} (2^(2k) - 1)

is provable. Right? 

I tested all  n < 10000  and found no counterexample.

But a proof seems far away..

Emmanuel. 

[Tomasz Ordowski]

Conjecture: (2n+1)!! divides Product_{k=1..n} (2^(2k) - 1). 

ChatGPT suggested an interesting p-adic/LTE approach that seems to reduce the problem to a comparison of Legendre-type valuations. I have not checked all the details yet, so I cannot claim it is a proof.

If anyone is curious, feel free to ask ChatGPT for the heuristic (or perhaps even a proof!). 

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[Ruud H.G. van Tol]

On 2026-07-10 20:22, Tomasz Ordowski wrote:
> [...]

> However, I believe that
> (2n+1)!! | Product_{k=1..n} (2^(2k) - 1)
> is provable. Right?

For n = 0, I presume this means: 1!! | 1, or as a pair: (1, 1).
More pairs: (3, 3), (15, 45), (105, 2835), ...

The division results as a sequence:
[1, 1, 3, 27, 765, 71145, 22410675, 24476939235, ...]

Their maximum factor that is congruent to {-1, 1} mod 2^n:
[1, 1, 3, 3, 17, 31, 31, 127, 257, 257, 257, 257, 257, 8191, ...]


Some exploitable patterns:

- - - - - - -

A000984: binomial(2*n, n) = (2*n)! / n! / n!.
[1, 2, 6, 20, 70, 252, 924, 3432, 12870, ...]

A000680: (2*n)! / 2^n = (2*n) * (2*n-1)! / 2^n.
[1, 1, 6, 90, 2520, 113400, 7484400, ...]

- - - - - - -

(A) For n >= 1, (2*n+1)!! = (2*n+1) * (2*n-1)!!
= (2*n+1) * (2*n)! / n! / 2^n
= (2*n+1) * binomial(2*n, n) * n! / 2^n.

- - - - - - -

(B) Product_{k=1..n} (2^(2k) - 1)
= Product_{k=1..n} (2^k - 1) * (2^k + 1)
= Product_{k=1..n} (2^k - 1) * Product_{k=1..n} (2^k + 1), which
trivially divides n!.

- - - - - - -

B / A = {1, 3, 27, 765, 71145, 22410675, 24476939235, ...}
which seems to be either divisible by (2^n-1) or (2^n+1), see above.

-- Ruud

[Tomasz Ordowski]

Yes, thanks, but still no proof that

 Product_{k=1..n} (4^k-1)/(2k+1) 

is an integer for every natural n. 

Ask ChatGPT (he knows something). 

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[Geoffrey Caveney]

The following observation may be useful:

For any odd prime p, 2^(p-1) == 1 mod p.

Thus, in the critical cases where 2n+1 = p, it follows that  2^(2n) = 2^(p-1) == 1 mod p == 1 mod 2n+1.

Since 2^(2n) == 1 mod 2n+1, then 2^(2n) - 1 == 0 mod 2n+1. Thus, 2^(2n) - 1 is divisible by 2n+1 in the cases where 2n+1 is prime.

Geoffrey

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[Tomasz Ordowski]

Yes, this obviously follows from Fermat's Little Theorem, but it's worth examining carefully the sequence of denominators of the quotient (4^n-1)/(2n+1). Such a sequence does not exist in the OEIS, and the closest one is A184726 (with a different offset).

See https://oeis.org/A184726 

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[Tomasz Ordowski]

PS. Proof: (2n+1)!! | A171080(n) | Product_{k=1..n} (4^k-1). Right? 

Note that A171080(n) = Product_{k=1..n} Den(B_{2k} / 2). 

See https://oeis.org/A171080

Cf. https://oeis.org/A091137

https://oeis.org/A363596 

[Ruud H.G. van Tol]

On 2026-07-13 23:03, Tomasz Ordowski wrote:
> PS. Proof: (2n+1)!! | A171080(n) | Product_{k=1..n} (4^k-1). Right?

The floor()s look troublesome to me.
That is also why I pointed at the +-1 dependencies.

-- Ruud

[Max Alekseyev]

On Fri, Jul 10, 2026 at 2:22 PM Tomasz Ordowski <tomaszo...@gmail.com> wrote:

However, I believe that 

(2n+1)!! | Product_{k=1..n} (2^(2k) - 1)

is provable. Right? 

Here is a proof. 

(2n+1)!! is composed of odd primes. Since (2n+1)!!=(2n+1)!/n!/2^n, for any odd prime p, we have 

v_p( (2n+1)!! ) = Sum_{k>=1} [(2n+1)/p^k] - [n/p^k].

On the other hand, since the multiplicative order ord_{p^k}(2) divides phi(p^k) = (p-1)p^(k-1) for any k>=1, we have

v_p( Product_{k=1..n} (2^(2k) - 1) ) >= Sum_{k>=1} [n / ((p-1)/2) / p^(k-1)] >= Sum_{k>=1} [2n/p^k].

Note that unless 2n+1=p^k, we have  [(2n+1)/p^k] - [n/p^k] <= [(2n+1)/p^k] = [2n/p^k].

In the case, 2n+1=p^e with e>1, we have

Sum_{k>=1} [(2n+1)/p^k] - [n/p^k] = Sum_{k>=1} [(2n+1)/p^k] -  Sum_{k>=1} [n/p^k]

<= (Sum_{k>=1} [(2n+1)/p^k]) - 1 = Sum_{k>=1} [2n/p^k],

where "-1" compensates at k=e.

In the remaining case 2n+1=p, we verify directly that

v_p( (2n+1)!! ) = 1 <= v_p( Product_{k=1..n} (2^(2k) - 1) ).

Hence, (2n+1)!! does divide Product_{k=1..n} (2^(2k) - 1).

Regards,

Max 

[Tomasz Ordowski]

Max, 

nice proof, 

well done!

Tom 

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[Tomasz Ordowski]

PS. Perhaps the stronger divisibility is also provable: 

(2n+1)!! | A171080(n) | Product_{k=1..n} (2^(2k) - 1),
where A171080(n) = Product_{k=1..n} Den(B_{2k})/2,
and occurs Den(B_{2k})/2 | 2^(2k) - 1. 

See https://oeis.org/A171080
Cf. https://oeis.org/A091137

and https://oeis.org/A363596 

[Tomasz Ordowski]

SUPPLEMENT to PS.

Max's proof seems to extend to the stronger divisibility

(2n+1)!! | A171080(n),

where

A171080(n) = Product_{k=1..n} Den(B_{2k})/2

= Product_{3<=p<=2n+1} p^floor(2n/(p-1))

(by the von Staudt-Clausen theorem).

Indeed,

v_p((2n+1)!!) = Sum_{k>=1} (floor((2n+1)/p^k) - floor(n/p^k))
<= Sum_{k>=1} floor(2n/p^k),

where the exceptional case 2n+1 = p^e is handled exactly as in Max's proof.

Finally, by Legendre's identity,

Sum_{k>=1} floor(2n/p^k) = (2n - s_p(2n))/(p-1) <= floor(2n/(p-1)),

hence v_p((2n+1)!!) <= v_p(A171080(n)) for every odd prime p, proving

(2n+1)!! | A171080(n).

Thus

(2n+1)!! | A171080(n) | Product_{k=1..n} (2^(2k)-1).

All right? 

Tom 

[Arup Kumar]

Hi Max,

Thank you for your valuable insight and the proof regarding the divisibility of the product of Mersenne-related terms by (2n+1)!!.

Building upon that, I believe the stronger divisibility relation involving A171080(n) is also provable:

(2n+1)!! \mid A171080(n) \mid \prod_{k=1}^{n} (2^{2k} - 1)

where A171080(n) = \prod_{k=1}^{n} \frac{\text{den}(B_{2k})}{2}.

The logic relies on the Clausen-von Staudt theorem, which states that \text{den}(B_{2k}) = \prod_{(p-1) \mid 2k} p. Since the prime factors of the denominators of Bernoulli numbers are exactly those primes p for which (p-1) \mid 2k, they are naturally embedded within the prime structure of (2^{2k}-1) (as p \mid 2^{p-1}-1 \implies p \mid 2^{2k}-1 when (p-1) \mid 2k).

This seems consistent with the properties discussed in A091137 and A363596. I would appreciate your thoughts on this extension.

Best regards,

Arup Kumar 

ORCID: 0009-0006-1019-8403

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