[seqfan] Re: (2p-1)!/(p-1)!^2 == p (mod p^4)
Tomasz Ordowski
tomaszordowski at gmail.com
Fri Jul 27 17:17:05 CEST 2018
P.S. Supplement.
The conjecture:
For n > 3, a(n-1) == 1 (mod n^3) if and only if n is prime,
where a(n-1) = binomial(2n-1,n) = A001700(n-1).
See https://oeis.org/A001700
The question is how to prove it?
Thomas
2018-07-27 14:11 GMT+02:00 Tomasz Ordowski <tomaszordowski at gmail.com>:
> Dear SeqFan,
>
> Conjecture: For n > 3, binomial(2n-1,n) == 1 (mod n^3) if and only if n is
> prime.
>
> Equivalently: (2p-1)!/(p-1)!^2 == p (mod p^4) only for all primes p > 3.
> No proof.
>
> Thus a(p-1) == p (mod p^4) for every prime p > 3, where a(n) = A002457(n).
>
> See https://oeis.org/A002457
>
> Question: is a proof of this conjecture known?
>
> Does this result from Wilson's theorem?
>
> Best regards,
>
> Thomas
>
>
>
>
>
> <#m_2849649430454898626_m_1360274943545487338_DAB4FAD8-2DD7-40BB-A1B8-4E2AA1F9FDF2>
>
More information about the SeqFan
mailing list