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

 

You received this message because you are subscribed to the Google Groups "SeqFan" group.
To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.
To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAF0qcNNc_SKeJXAwNz7zKBgw-HSpBATXkvkwpDUvDKja%2BXjwJQ%40mail.gmail.com.

[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  

--

You received this message because you are subscribed to the Google Groups "SeqFan" group.
To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/c35d955b-5a4c-4c62-88db-e4ecd958ebe5n%40googlegroups.com.

[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!). 

--

You received this message because you are subscribed to the Google Groups "SeqFan" group.
To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAEiteAMB8dMyTxievRce_h4RR%3DaH_qRPM2WnMNC-T%2BMwBEsyiQ%40mail.gmail.com.

[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). 

--
You received this message because you are subscribed to the Google Groups "SeqFan" group.
To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/ab02101c-503f-4a8a-a995-8b27fea06f1b%40isolution.nl.

[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

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAF0qcNNO5rePB_hRSzY98oarM3vkJyTYEkY3EdqavUp4_YxUvA%40mail.gmail.com.

[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 

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAFhKorUL462pyYYGGEjCcWzEnUdmY3rPvnUy3E5rCxN9diobRA%40mail.gmail.com.

[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 

--

You received this message because you are subscribed to the Google Groups "SeqFan" group.
To unsubscribe from this group and stop receiving emails from it, send an email to seqfan+un...@googlegroups.com.

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAJkPp5O-AXUSKe1K5CLbK7f6vA%2BFHR%3D-hxgbfbkvzEUmAChOog%40mail.gmail.com.

[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

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAF0qcNOvTJ9PZzh9kbh9OrDFxAh%3D3mhEYmm7oKgogdQ9KAdqHQ%40mail.gmail.com.

[Tomasz Ordowski]

Dear Arup!

Well, that's exactly what I used in 

my proof of this stronger divisibility, 

modeled after Max's. Didn't you notice?

Best, 

Thomas Ordowski  

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CABL08hZd1SqeXwGzq7ck0D63DD4XoBSrGy9yyb1v5n11qkxZ2Q%40mail.gmail.com.

[Arup Kumar]

Dear ,

I am writing to share a formal derivation regarding the divisibility relationship between the double factorial (2n+1)!! and the product of Bernoulli denominators, A171080(n).

### Theorem

For all n >= 1, (2n+1)!! | A171080(n), where:

 * (2n+1)!! = Product_{k=0 to n} (2k+1)

 * A171080(n) = Product_{k=1 to n} (Den(B_{2k}) / 2) = Product_{p <= 2n+1} p^floor(2n / (p-1)) (by the von Staudt-Clausen theorem).

### Proof

 1. Prime Mapping and Density

   For any odd prime p, let E_p(A) be the p-adic valuation of A171080(n). According to the von Staudt-Clausen theorem, p divides Den(B_{2k}) if and only if (p-1) divides 2k. In the range k in {1, ..., n}, the number of such k is the number of multiples of (p-1)/2 in {1, ..., n}, which is exactly floor(n / ((p-1)/2)) = floor(2n / (p-1)). Thus, E_p(A) = floor(2n / (p-1)).

 2. Valuation of the Double Factorial

   Using the identity (2n+1)!! = (2n+1)! / (2^n * n!), we apply Legendre’s formula for p > 2:

   v_p((2n+1)!!) = v_p((2n+1)!) - v_p(n!) = Sum_{j >= 1} (floor((2n+1) / p^j) - floor(n / p^j)).

 3. Comparison and Conclusion

   The term f_j = floor((2n+1) / p^j) - floor(n / p^j) represents the count of integers in the interval (n, 2n+1] divisible by p^j. Summing over all j >= 1, we observe that the total p-adic density is bounded by the sum of the geometric series Sum_{j >= 1} (2n / p^j) = 2n / (p-1).

Since v_p((2n+1)!!) <= floor(2n / (p-1)) and E_p(A) = floor(2n / (p-1)), it follows that for every odd prime p:

v_p((2n+1)!!) <= v_p(A171080(n)).

This confirms the divisibility (2n+1)!! | A171080(n), and by extension, the divisibility of the product of Mersenne-type expressions.

I would welcome your feedback on this density-based approach to the divisibility chain.

Best regards,

Arup Kumar

ORCID: 0009-0006-1019-8403

To view this discussion visit https://groups.google.com/d/msgid/seqfan/CAF0qcNPSTyhWpZtXVsMkB5_mgPSg2%3DjpVQB%3D6Fs1kA5AkNdoQg%40mail.gmail.com.