[seqfan] Does the equation D_{n-1} = 2n only have one solution?

[Tomasz Ordowski]

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