# About the number of the elements of a set related with binomial coefficients

For $N\in\mathbb N$, let $$P_l(N)=\# \{(n,m)|0\le n\le N, 0\le m\le n,\binom{n}{m}\not\equiv 0 \mod l\}.$$ Suppose that $\binom{n}{0}=1$ for $n\ge 0$ and that $\# S$ represents the number of the elements of a set $S$.

Then, here are my questions.

Question 1 : Find $\lim_{N\to\infty}\log_N P_6(N)$.

Question 2 : Find $\lim_{N\to\infty}\log_N P_l(N)$ for $l={p_1}^{m_1}{p_2}^{m_2}\cdots {p_s}^{m_s}\ (p_1\lt p_2\lt \cdots\lt p_s)$.

Motivation : I've found that $\lim_{N\to\infty}\log_N P_p(N)=\log_p \frac{p(p+1)}{2}$ for any prime number $p$. However, I'm facing difficulty for non-prime-number cases. Can anyone help?

Remark : This question has been asked previously on math.SE without receiving any answers.

• You may find something of interest in Granville's survey article on arithmetical properties of binomial coefficients, freely available at dms.umontreal.ca/~andrew/Binomial – Gerry Myerson Oct 27 '13 at 22:41

${a \choose b} \not \equiv 0 \mod 6$ iff either ${a \choose b} \not \equiv 0 \mod 2$ or ${a \choose b } \not \equiv 0 \mod 3$. So, $$P_3(N) \le P_6(N) \le P_2(N) + P_3(N).$$
Since $P_2(N) = \Theta(N^{\log_2 3}) = \Theta(N^{1.585})$ is asymptotically negligible compared with $P_3(N) = \Theta(N^{\log_3 6}) = \Theta(N^{1.631})$,
$$P_6(N) \sim P_3(N).$$
In general, for square-free $\ell$ with largest prime factor $q$, $P_\ell(N) \sim P_q(N).$
I haven't checked the details, but I expect that if the largest prime power dividing $\ell$ is $q = p^k$, then $P_\ell(N) \sim P_q(N)$ and $\lim_{N\to\infty} \log_N P_\ell(N) = \log_p \frac{p(p+1)}{2}$. Edit: Actually, $q$ should be the power of the largest prime dividing $\ell$, not the largest prime power. If $\ell = 144 = 2^4 3^2$ then for large $N$, most of the binomial coefficients in the first $N$ rows of Pascal's triangle not divisible by $144$ should be divisible by $3$ and $16$ but not $9$.
• I think I'll be able to prove that the answer for $(2)$ is $\log_{p_s}\frac{p_s(p_s+1)}{2}$ from your idea. Thank you so much! – mathlove Oct 28 '13 at 14:39