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?

#Combinatorics #Proofs #MathProblem #Math

Note by Ivan Stošić
7 years, 10 months ago

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14 votes

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Comments

Your conjecture is actually true. As illustrated in this article, the following holds v3(∑k=0n−1(2kk))=v3((2nn))+2v3(n)v_3\left (\sum_{k=0}^{n-1}\binom{2k}{k}\right)=v_3\left (\binom{2n}{n}\right)+2v_3(n)

where v3(n)v_3(n) is equal to the exponent of 33 in the prime factorization of nn.

Your result is a direct corollary of this fact.

Shyan Akmal - 7 years, 10 months ago

Does anybody have an elementary argument? Or an elementary introduction to pp-adic numbers? Could a combinatorial argument use the facts that 3k−1=2(1+3+32+⋯+3k−1)3^k -1 = 2(1+3+3^2+\cdots+3^{k-1}) and 32k=9k=(1+8)k=∑j(kj)8j3^{2k} = 9^k = (1+8)^k = \sum_j \binom{k}{j} 8^j ?

Eric Edwards - 7 years, 10 months ago
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