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