Divisibility of central binomial coefficients

Importance: Medium ✭✭

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Subject:	Number Theory
	» Combinatorial Number Theory

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Recomm. for undergrads: yes

Prize: $1,000 each

Posted	by:	maxal
	on:	August 5th, 2007

Problem (1) Prove that there exist infinitely many positive integers $n$ such that

$\gcd({2n\choose n}, 3\cdot 5\cdot 7) = 1.$

Problem (2) Prove that there exists only a finite number of positive integers $n$ such that

$\gcd({2n\choose n}, 3\cdot 5\cdot 7\cdot 11) = 1.$

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

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.

Sequence A030979: Numbers n such that C(2n,n) is not divisible by 3, 5 or 7.

Andrew Granville. "The Arithmetic Properties of Binomial Coefficients."

* indicates original appearance(s) of problem.

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

On January 18th, 2012 Anonymous says:

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).

Open Problem Garden

Divisibility of central binomial coefficients

Bibliography

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

Reference for problem (2)

Problem 1 solved?

Does not look like it.

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