A conjecture in A027423 is stated as a fact in A068499

[Sela Fried]

In the comments to A027423 it reads:

It appears that a(n+1)=2*a(n) if n is in A068499. - Benoit Cloitre, Sep 07 2002

but in the comments to A068499 (authored by Cloitre) it reads:

Also n such that tau((n+1)!) = 2* tau(n!)

which is exactly the conjecture...

so, is it a conjecture or a fact?

 

[Neil Sloane]

I will forward that to Benoit Cloitre (I don't know if he is a member of this list or not)

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

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[Jonas Karlsson]

Apart from small values, the only way m+1 can fail to divide m! is if m+1 is prime, which causes tau((m+1)!) = tau(m!)*tau(m+1) = 2*tau(m!). So the observation is correct.

J

[Neil Sloane]

Jonas,  That's great.  Can you make the appropriate corrections to thos two sequences?

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

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[Jonas Karlsson]

Can do!

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[jp allouche math]

By the way, if one does not see immediately why "Apart from small values... if m+1 is prime", one can
look at Wikipedia under "Wilson's theorem". Look for the setnce that begins with "In fact, more is true"

best
jean-paul

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[Neil Sloane]

I forwarded the first message to Benoit Cloitre (benoi...@yahoo.fr).   He tells me he is going to join the SeqFan list after he changes his email address.  He asked me to post two messages on his behalf.  Here they are.

#1

Benoit Cloitre

Oct 22, 2025, 6:21 PM (14 hours ago)

to me

Dear Neil,

Thank you for writing. I will consider Seqfan subscription!

I would like to clarify the current state of comments in OEIS regarding sequences A068499 and A027423, and explain the complete mathematical result I have now proved.

CURRENT STATE OF MY OEIS COMMENTS:

For A068499 (numbers m such that m! is not divisible by m+1): The comment states "Also n such that tau((n+1)!) = 2* tau(n!)" (which is not mine I think). This phrasing is ambiguous: it reads like a complete characterization ("all n such that..."), but it could mean only one direction (if n is in this sequence, then tau((n+1)!) = 2tau(n!)), which would have been clearer if written as "Also, if n is in this sequence then tau((n+1)!) = 2tau(n!)".

For A027423 (number of divisors of n!): My 2002 comment states "It appears that a(n+1) = 2*a(n) if n is in A068499." This explicitly conjectures only one direction (sufficiency).

MY COMPLETE RESULT (2025):

I have now proved the full equivalence (if and only if): tau((n+1)!) = 2tau(n!) if and only if n is in A068499, equivalently: tau((n+1)!) = 2tau(n!) if and only if n = 3 or n+1 is prime.

THE TWO DIRECTIONS:

1. Sufficiency: If n is in A068499, then tau((n+1)!) = 2*tau(n!). This direction is easy: Wilson's theorem tells us that n in A068499 means n+1 is prime or n=3, and the result follows from multiplicativity of tau (for primes) or direct computation (for n=3).
2. Necessity: If tau((n+1)!) = 2*tau(n!), then n is in A068499. This is the nontrivial direction.

THE NONTRIVIAL PART (necessity):

Wilson's theorem gives a dichotomy: either (n+1) does not divide n! (Case 1: n+1 prime or n=3, which gives the desired conclusion), or (n+1) divides n! (Case 2: n+1 composite and n+1 not equal to 4, which must be ruled out).

In Case 2, one must prove that the equation tau((n+1)!) = 2*tau(n!) cannot hold. The proof uses Legendre's formula for prime valuations in factorials and an asymptotic argument (splitting by largest prime factor) showing the product is strictly less than 2 for all composite n+1 not equal to 4.

This completes my 2002 conjecture into a full theorem (if and only if) and clarifies the ambiguous comment on A068499.

I have written a complete proof in a short note that I can provide.

Best regards,

Benoit

#2 

Benoit Cloitre

2:19 AM (6 hours ago)

to me

Hi Neil,

I’m not on seqfans yet, since I plan to change my OEIS email soon

(benoi...@yahoo.fr -->benoit....@proton.me). 

Here a note with the proof.

You’re welcome to share the contents of my previous email and the note on seqfan.

Best regards,

Benoit

#3  What Benoit sent me is a pdf file.  I will try to attach it here

[Jonas Karlsson]

Great! I submitted two comments, one of which claimed the converse. I did in fact go through the reasoning that Benoit wrote up but my comment is rather terse. Anyone feel free to delete it - or I suppose the pdf with the proof could be added? 

J

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