# Divisibility of central binomial coefficients

\begin{problem}[1] Prove that there exist infinitely many positive integers $n$ such that $$\gcd({2n\choose n}, 3\cdot 5\cdot 7) = 1.$$ \end{problem}

\begin{problem}[2] Prove that there exists only a finite number of positive integers $n$ such that $$\gcd({2n\choose n}, 3\cdot 5\cdot 7\cdot 11) = 1.$$ \end{problem}

% You may use many features of TeX, such as % arbitrary math (between $...$ and $$...$$) % \begin{theorem}...\end{theorem} environment, also works for question, problem, conjecture, ... % % Our special features: % Links to wikipedia: \Def {mathematics} or \Def[coloring]{Graph_coloring} % General web links: \href [The On-Line Encyclopedia of Integer Sequences]{http://www.research.att.com/~njas/sequences/}

The binomial coefficient ${2n\choose n}$ is not divisible by prime $p$ iff all the base-$p$ digits of $n$ are smaller than $\frac{p}{2}.$

It has been conjectured that 1, 2, 10, 3159, and 3160 are the only positive numbers for which $\gcd({2n\choose n}, 3\cdot 5\cdot 7\cdot 11) = 1$ holds.

## Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

\href[P. Erdos, R.L. Graham, I.Z. Ruzsa and E.G. Straus "On the Prime Factor of $2n \choose n$." Math. Comp. 29 (1975), 83-92.]{http://www.math.ucsd.edu/~sbutler/ron/75_03_prime_factors.pdf}

\href[Sequence A030979: Numbers n such that C(2n,n) is not divisible by 3, 5 or 7.]{https://oeis.org/A030979}

\href[Andrew Granville. "The Arithmetic Properties of Binomial Coefficients."]{http://www.cecm.sfu.ca/organics/papers/granville/index.html}

* indicates original appearance(s) of problem.

### 2, 10, and 3159 are not valid for problem (2)

In the initial comment five values are stated to be not divisible by 3, 5, 7, and 11. That is true for 1 and 3160 but not for 2, 10, and 3159:

The last base-3-digit of 2 is 2. The last base-11-digit of 10 is 10. The last base-5-digit 3159 is 4.

Or check these facts:

${{2 \times 2} \choose 2} = 6 = 3 \times 2$

${{2 \times 10} \choose 10} = 184756 = 11 \times 16796$

2 and 3159 are not in the sequence A030979 (mentioned in the bibliography).

### Reference for problem (2)

It appears that the right reference for question 2 should be \href[A151750]{http://oeis.org/A151750} instead.

### Problem 1 solved?

Looks like Problem 1 was solved by http://arxiv.org/PS_cache/arxiv/pdf/1010/1010.3070v1.pdf

## It seems Problem (1) was

It seems Problem (1) was solved last year, see http://arxiv.org/abs/1010.3070

How about Problem 2?