[seqfan] Re: Does the equation D_{n-1} = 2n only have one solution?
Ami Eldar
amiram.eldar at gmail.com
Thu Jun 10 10:46:14 CEST 2021
I searched the 4276 terms in the b-file of A258801 but did not find there
any number n such that D_{n-1} = 2pn.
Best,
Amiram
On Tue, Jun 8, 2021 at 4:07 PM Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:
> Dear readers!
>
> Let D_k be the denominator of Bernoulli number B_k.
> For odd n > 1, D_{n-1} = Product_{p prime, p-1|n-1} p.
> Problem: Are there numbers n > 3 such that D_{n-1} = 2n ?
> If so, they are Carmichael numbers n > 2^64 divisible by 3.
> Note that, if p is a prime, then D_{p-1} = 2p only for p = 3.
>
> Primes p such that D_{p-1} = 6p are https://oeis.org/A092307
> Composites n such that D_{n-1} = 6n are Carmichael numbers
> 310049210890163447, 18220439770979212619, ...
> Found by Amiram Eldar.
>
> Are there numbers n such that D_{n-1} = 2pn with prime p > 3 ?
> If so, such n must be a Carmichael number divisible by 3.
> Search database https://oeis.org/A258801
>
> Best regards,
>
> Thomas
> ____________________
> https://www.bernoulli.org/~bk/denombneqn.pdf
> https://math.dartmouth.edu/~carlp/pomwag060121.pdf
>
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