[seqfan] Re: A064169

[Max Alekseyev]

Copying my answer I sent to Thomas earlier today:

Wolstenholme's theorem gives
N(p-2)/D(p-2) + 1/(p-1) == 0 (mod p^2),
that is
(p-1)*N(p-2) + D(p-2) == 0 (mod p^2).
Modulo p this reduces to
N(p-2) == D(p-2) (mod p).

Regards,
Max

On Thu, Aug 1, 2019 at 10:24 AM Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:

> Dear SeqFans,
>
> I have an interesting conjecture and a related question.
>
> Let H(n) = 1/1 + 1/2 + ... + 1/n is the n-th Harmonic number.
> Let N(n) = Numerator(H(n)) and let D(n) = Denominator(H(n)).
>
> The conjecture: For n > 2,
> N(n-2) == D(n-2) (mod n) if and only f n is a prime.
>
> The question: Is, by Wolstenholme's theorem,
> if p is an odd prime, then N(p-2) == D(p-2) (mod p) ?
>
> Maybe someone knows the answer.
>
> Best regards,
>
> Thomas
> _________________________
> https://arxiv.org/abs/1111.3057
> https://oeis.org/history/view?seq=A064169&v=61
> https://en.wikipedia.org/wiki/Wolstenholme%27s_theorem
> http://mathworld.wolfram.com/WolstenholmesTheorem.html
>
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>