Erdős Problem #1093

declared status 'open'. Formalized: yes. For $n\geq 2k$ we define the deficiency of $\binom{n}{k}$ as follows. If $\binom{n}{k}$ is divisible by a prime $p\leq k$ then the deficiency is undefined. Otherwise, the deficiency is the number of $0\leq i<k$ such that $n-i$ is $k$-smooth, that is, divisible only by primes $\leq k$. Are there infinitely many binomial coefficients with deficiency $1$? Are there only finitely many with deficiency $>1$? Current best: In [ELS93] they prove that if the deficiency exists and is $\geq 1$ then $n\ll 2^k\sqrt{k}$. The following have deficiency $1$ (there are $58$ examples with $n\leq 10^5$):$$\binom{7}{3},\binom{13}{4},\binom{14}{4},\binom{23}{5},\binom{62}{6},\binom{94}{10},\binom{95}{10}.$$The examples which follow are the only known examples with deficiency $>1$. Prize: no. Tags: binomial coefficients, number theory.

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