[seqfan] Does the equation D_{n-1} = 2n only have one solution?
Tomasz Ordowski
tomaszordowski at gmail.com
Sun Jun 6 07:40:18 CEST 2021
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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