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?

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TopNewestYour 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. – Shyan Akmal · 4 years ago

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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 \(3^{2k} = 9^k = (1+8)^k = \sum_j \binom{k}{j} 8^j\) ? – Eric Edwards · 4 years ago

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