# Divisibility of central binomial coefficients

 Importance: Medium ✭✭
 Author(s): Graham, Ronald L.
 Subject: Number Theory » Combinatorial Number Theory
 Keywords:
 Prize: $1,000 each  Posted by: maxal on: August 5th, 2007 \begin{problem}[1] Prove that there exist infinitely many positive integers$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. ### It seems Problem (1) was It seems Problem (1) was solved last year, see http://arxiv.org/abs/1010.3070 How about Problem 2? ### 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

### Does not look like it.

Does not look like it.