[seqfan] Re: (b^k-1)/(b-1) == 1 (mod k)

[David Wilson]

> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Tomasz
> Ordowski
> Sent: Tuesday, April 24, 2018 3:43 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] (b^k-1)/(b-1) == 1 (mod k)
> 
> Dear SeqFan!
> 
> Let a(n) be the smallest composite k such that
> 
>          b^k == b (mod (b-1)k)
> 
> for all integer bases 2 <= b <= n = 2, 3, 4, 5, ...

This looks to be A302588, which you recently submitted.
If so, the attached b-file b302588.txt extends the sequence out to a(211).

> Conjecture: for n > 2, a(n) is a Carmichael number. Yes?

I wouldn't be surprised if for any n >= 2, you can find some Carmichael number k = c(n) satisfying b^k == b (mod (b-1)k) for 2 <= b <= n.
This proof may be conditional on some reasonable conjecture about Carmichael numbers, e.g., the existence of an infinite number of Carmichael numbers of some specific form.
Such a proof would (perhaps conditionally) prove that a(n) exists for all n >= 2.
My intuition is that as n grows, stochastic considerations (the large number of congruences that must be satisfied) will strongly favor a(n) being a Carmichael number. You might be able to cob together a stochastic argument that a(n) is almost certainly Carmichael for large n, I doubt you find a proof.  The problem seems similar to trying to show that any sufficiently large power of 2 includes all the base-10 digits.  From a stochastic standpoint, large powers of two almost certainly include all the digits, but you can't prove it.

> Can also consider the congruence b^(k-1) == 1 (mod (b-1)k).

I created a b-file for that as well, the result coincides with A271221 as far as it goes (and extends the b-file).
I suspect your definition is equivalent to A271221, and this is likely provable.

I also suspect that your observation that A302588(n) = A271221(n - 1) for n >= 3 is in indeed correct, and likely provable.

> Maybe someone will be interested in these sequences.
> 
> Please give me some initial terms.
> 
> Best regards,
> 
> Thomas
> 
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