# Sum of central binomial coefficients

I was summing up some central binomial coefficients - those of the form $$\binom{2n}{n}$$ when I noticed that $$3^{2k}$$ divides $$\sum\limits _{n=0}^{3^k-1}\binom{2n}{n}$$ and is the highest power of 3 to do so i.e. $$3^{2k+1}$$ does not.

I have verified this statement for k up to 6. Can anyone find a counterexample, or have an idea how to prove this?

Note by Ivan Stošić
4 years, 12 months ago

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Your conjecture is actually true. As illustrated in this article, the following holds $v_3\left (\sum_{k=0}^{n-1}\binom{2k}{k}\right)=v_3\left (\binom{2n}{n}\right)+2v_3(n)$

where $$v_3(n)$$ is equal to the exponent of $$3$$ in the prime factorization of $$n$$.

Your result is a direct corollary of this fact.

- 4 years, 12 months ago

Does anybody have an elementary argument? Or an elementary introduction to $$p$$-adic numbers? Could a combinatorial argument use the facts that $$3^k -1 = 2(1+3+3^2+\cdots+3^{k-1})$$ and $$3^{2k} = 9^k = (1+8)^k = \sum_j \binom{k}{j} 8^j$$ ?

- 4 years, 11 months ago