Re: [PrimeNumbers] Binomial and prime congruence
- --- Sebastian Martin <sebi_sebi@...> wrote:
> Hello all:It's a 3-line induction proof.
> Can anyone prove the following congruence?
> Binomial(m,p)=Floor[m/p] (mod p)
> for all m integer and for all p prime number.
> Corollary:Not that I know of. I know that the power of each prime that divides a binomial
> If pn and p(n+1) are consecutive pime numbers
> Binomial[p(n+1),pn]=1 (mod pn)
> (it is easy, consider Bertrand postulate)
> There are any previous reference?
expression is one that has been mentioned in the past. The expressions
discovered are useful for fast computation of binomials.
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