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The central binomial coefficients are those integers $$\binom{2n}n=\frac{(2n)!}{(n!)^2}\ \ \ (n=0,1,2,\ldots).$$

QUESTION: Does each odd prime $p$ have a primitive root $g<p$ which is the sum of two central binomial coefficients?

I conjecture that the answer is yes. For example, $$\binom{2\times3}3+\binom{2\times3}3=40\ \ \text{and}\ \ \ \binom{2\times1}1+\binom{2\times4}4=72$$ are primitive roots modulo the prime $109$. For the number of pairs $\{k,m\}$ ($0\leqslant k\leqslant m$) with $\binom{2k}k+\binom{2m}m$ not only smaller than the $n$th prime $p_n$ but also a primitive root modulo $p_n$, one may visit http://oeis.org/A305030. I have verified that each odd prime $p<10^9$ has a primitive root $g<p$ which is the sum of two central binomial coefficients.

Recall that the Catalan numbers are given by $$C_n=\frac1{n+1}\binom{2n}n=\binom{2n}n-\binom{2n}{n+1}\ \ (n=0,1,2,\ldots).$$ I also guess that every odd prime $p$ has a primitive root $g<p$ which is the sum of two Catalan numbers.

Any comments are welcome!

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  • $\begingroup$ "the sum of two central binomial coefficients" - you mean the sum modulo $p$, or the genuine sum as a real number? $\endgroup$ – Fedor Petrov May 25 '18 at 7:47
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    $\begingroup$ @Fedor I think I have already clarified that. $g=\binom{2k}k+\binom{2m}m$ is an integer smaller than $p$ and it is also a primitive root modulo $p$. $\endgroup$ – Zhi-Wei Sun May 25 '18 at 9:18

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