For any sequence of positive integers $\{ a_i \}_{i \ge 1}$ we can define the generalized binomial coefficients $\binom{n}{k}_{a}$ as follows: $$m!_a = a_1 a_2 \cdots a_m, \binom{n}{k}_a = \frac{n!_a}{k!_a (n-k)!_a}$$

When $\{ a_i \}$ is a strong divisibility sequence, i.e. $\gcd (a_n, a_m) = a_{\gcd(n,m)}$, it can be shown that those coefficients are always integral, see this paper by Ward.

The problem is that this is only a sufficient condition, and I am looking for a neccesary condition.

Why is it only sufficient?

  1. Because it proves a stronger reuslt, namely that $\binom{n+1}{k+1}_a$ can be written as a linear integral combination of $\binom{n}{k+1}_a, \binom{n}{k}_a$.

  2. There are examples that don't satisfy the conditions but give integral coefficients, for example: $a_n = \binom{n+c}{c+1}$ for $c \ge 1$, see this thesis. Note that $(a_2, a_3) = (c+2, \frac{(c+3)(c+2)}{2})\ge \frac{c+2}{2} > 1$, yet $a_{(2,3)}=a_1 = 1$.

What have I come up with?

  1. If the binomial coefficients are integral, we have: $a_{n+1} \cdot \cdots \cdot a_{n+k}$ is divisible by $a_{1} \cdots a_{k}$. For any prime $p$ define the sequence $\{ b_{n,p} = v_p(a_n) \}_{n \ge 1}$. We have the condition that any sum of $k$ consecutive integers in the sequence is at least as large as the sum of the first $k$ integers in the sequence.

  2. We also have the condition that for any $n$, $v_p(a_n) \neq 0$ for finitely many primes $p$.

What do I want?

  1. A full compact characterization of sequences giving rise to integral binomial coefficients.

  2. Examples of interesting such sequences which are not strong divisibility sequences.

  • $\begingroup$ Did you make any headway regarding this problem? $\endgroup$ – Henry Sep 18 '18 at 20:19
  • $\begingroup$ @Henry No, I did not. $\endgroup$ – Ofir Sep 18 '18 at 21:44

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