[seqfan] Numbers n such that D_{n-1} = 2(n-2)n

[Tomasz Ordowski]

Dear readers!

Let D_k be the denominator of Bernoulli number B_k.
For k > 0, D_{2k} = Product_{p prime, p-1|2k} p.

Let's define:
Primes p such that D_{p-1} = 2(p-2)p.
Are these only the Fermat primes?
Are there Carmichael numbers m
such that D_{m-1} = 2(m-2)m ?

Conjecture:
D_{n-1} = 2(n-2)n if and only if n is a known Fermat prime.

Note: Let p = 2^{2^4}+1 = 65537, the last known Fermat prime.
If there exists a next Fermat prime q, it satisfies another equation,
namely D_{q-1} = 2(p-2)pq = 2(2^{2^5}-1)q = 2(3*5*17*257*65537)q.

Best regards,

Thomas
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https://oeis.org/A002997
https://oeis.org/A019434
https://oeis.org/A027642