[seqfan] The Bernoulli numbers harmony

[Tomasz Ordowski]

Dear readers,

I have another observation regarding Bernoulli numbers.

Let F(n) = Sum_{k=1..n} k B_{k-1} = N(n)/D(n), for n > 0.
Data: 1/1, 0/1, 1/2, 1/2, 1/3, 1/3, 1/2, 1/2, 1/5, 1/5, 31/30, ...
I noticed that if p is an odd prime, then p | N(p-2) + 2 D(p-2).
Conjecture: F(n-2) == - 2 (mod n) if and only if n is prime.
   In other words (without the duplicate terms).
Let S(n) = Sum_{k=1..n} (2k+1) B_{2k} = N(n)/D(n), for n > 0.
Data: 1/2, 1/3, 1/2, 1/5, 31/30, -79/35, 1067/70, -11059/105, ...
I noticed that if p = 2n+1 > 3 is prime, then p | N(n-1) + 2 D(n-1).
Conjecture (maybe it is well known to someone, but not to me):
   S(n-1) == -2 (mod 2n+1) if and only if p = 2n+1 is prime.
Question: is it equivalent to the Agoh-Giuga conjecture?
https://en.wikipedia.org/wiki/Agoh%E2%80%93Giuga_conjecture
Note: p B_{p-1} == -1 (mod p), i.e. p N_{p-1} == - D_{p-1} (mod p^2),
because p | D_{p-1}, where the Bernoulli number B_k = N_k / D_k.

I have recently written about the sequence
   F(n) = Sum_{k=1..n} (B_{k-1}+1/k) of fractions
2/1, 2/1, 5/2, 11/4, 35/12, 37/12, 13/4, 27/8, 1243/360, ...
(but no response) and put forward the strong conjecture:
For p > 3, F(p-2) == 0 (mod n) if and only if p is prime.
Equivalently, F(p-1) == -1 (mod p), by the equality
F(2n) = F(2n-1) + 1/(2n), for n > 1.

Best regards,

Thomas Ordowski
_______________________
https://en.wikipedia.org/wiki/Wolstenholme%27s_theorem
https://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem
https://en.wikipedia.org/wiki/Kummer%27s_congruence
All of this can come in handy.