I saw today in functions.wolfram.com a congruence for the prime counting function which says $\binom {2prime(k)-1} {prime(k)-1} \pmod{prime(k)^3}=1$ (the third congruence at the bottom).
What does it actually say?
(prime(n) is the notation for $\pi(n)$)
I thought at the beginning that the congruence states
$\binom {2\pi(k)-1} {\pi(k)-1}\equiv 1 \pmod{(\pi(k))^3}$
which is clearly false (for $k=10$ for example $\binom {2\pi(10)-1}{\pi(10)-1}=\binom {7}{3} \equiv 1\pmod{(\pi(10))^3}$ is false.)
I would like if someone understands what is Wolfram saying, to give me a reference or a proof of the congruence.
Thank you very much!
It is not the prime counting function but the $k$-th prime.
Thus, it is asserted that $$ \binom{2p-1}{p-1} \equiv 1 \pmod{p^3} $$ for $p$ a prime at least $5$; there is a condition $k>2$ there so it only holds from the third prime on.
This congruence is known as Wolstenholme's theorem.