@GerhardPaseman thanks for the comment but I was looking for a more fine-grained answer: just "often starts as a sequence of ones" is not good enough for me.

Thanks a lot! A follow up question: has anyone studied the conditional distribution of the diagonal terms in the Smith normal form, given the diagonals on the Hermite normal form?

Thank you for the answer! It is kind of surprising to me that simple Taylor approximation is not too far away from this bound. The error is $K^n/n!$, which is roughly $(eK/n)^n / \sqrt{2\pi n}$. Well, still there is a $2^n$ factor difference.

Hi @Michael Griffin, I have some follow-up questions. I tried your formula for $n = 2$ but it gives me the zero polynomial. Did I mess up somewhere? The reason I got zero is I computed $a(m) = 0$ for all $0 \leq m \leq p$..

Thanks! I'll try to walk through your argument and see if I can reproduce your results:)