[seqfan] Re: A new look at the Lehmer's totient problem

[Tomasz Ordowski]

CORRECTION AND SUPPLEMENT:

(**) [...]

The probable number of such Carmichael numbers
is 0, 0, 1, 6, 7, 9, 9, 22, 28, 29, 31, 36, 36 for k = 1, 2, ..., 13.
For now, we have a database of Carmichaels n < 10^18.
It's worth search all n < 10^21.

T. Ordowski

niedz., 10 lut 2019 o 15:40 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

> Dear SeqFans!
>
> Let a(n) = phi(n) / gcd(phi(n), n-1).
>
> (*) Conjecture: a(n) > 2 for every composite n > 6.
>   Note: Checked only for Carmichael numbers n.
> Are there no other counterexamples, for sure?
>
> (**) Conjecture:
> For each natural number k,
> there are at most finitely many
> Carmichael numbers n such that a(n) <= k.
>
> (**) Question: Could this be true?
>
> The probable number of such Carmichaels
> is 0, 0, 1, 5, ? for k = 1, 2, 3, 4, 5.
> For now, we have a database of
> Carmichael numbers n < 10^18.
> It's worth search all n < 10^21.
>
> I am asking for comments.
>
> Best regards,
>
> Thomas Ordowski & Amiram Eldar
> ______________________
> (*) By Lehmer's Conjecture, a(n) > 1 for every composite n.
> https://en.wikipedia.org/wiki/Lehmer%27s_totient_problem
> http://mathworld.wolfram.com/LehmersTotientProblem.html
>
> P.S. Further. Let's define:
> Carmichael numbers n such that a(n) is prime.
> (**) Conjecture:
> There are only finitely many such numbers?
> Data:
> n = 1729,
> 3069196417,
> 23915494401,
> 1334063001601,
> 6767608320001,
> 33812972024833,
> 1584348087168001,
> 1602991137369601,
> 6166793784729601,
> ?
> a(n) = 3, 11, 421, 23, 43, 41, 11, 5, 43, ?
> No more found for n < 10^18. Search all n < 10^21.
>
> Thomas Ordowski (idea) &
> Amiram Eldar (implementation)
>
>
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