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Is there a simple formula for the following? $$f(n) = \binom{2n}{n} \pmod{n^3}$$

I know $f(n) = 2$ iff $n$ is prime and greater than $3$, but I don't know anything about composite numbers.

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    $\begingroup$ $f(n)=2$ iff $n$ is prime? Not true for $n=2$ or $3$. $\endgroup$ – Robert Israel Jan 11 '19 at 19:54
  • $\begingroup$ Apologies, should be primes greater than 3. $\endgroup$ – Ben Crossley Jan 11 '19 at 20:25
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    $\begingroup$ Might help to look for papers that generalize Lucas's theorem. $\endgroup$ – DanielV Jan 11 '19 at 21:40
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    $\begingroup$ I first observed this by using Lucas's theorem (though I didn't know it by name) if we take the squares of row n of binomial triangle and sum them this gives $\binom{2n}{n}$ Then by Lucas theorem if we take away 2 it must be divisible by $n^2$ For reasons I dont know this can be increased to $n^3$ for primes greater than 3. $\endgroup$ – Ben Crossley Jan 11 '19 at 23:06
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    $\begingroup$ For $p$ prime, $p>3$, ${2p \choose p} \equiv 2 \bmod p^3$ is just a reformulation of Wolstenholme Theorem. $\endgroup$ – René Gy Jan 12 '19 at 13:56

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