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](https://math.stackexchange.com/questions/531186/about-the-number-of-the-elements-of-a-set-related-with-binomial-coefficients) without receiving any answers.



About the number of the elements of a set related with binomial coefficients
${a \choose b} \not \equiv 0 \mod 6$ 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 $9$.


