In view of Chebyshev's approach to prime numbers, I would like to ask about the regularities and peculiarities of the two sequences $\ \beta(n)\ $ and $\ \gamma(n),\ $ which I define as follows:


* $\ \beta(n)\ $ is the smallest integer $\ b\ $ such that $\ \binom bn\ $ has at least $\ n\ $ different prime divisors;
* $\ \gamma(n)\ $ is the smallest integer $\ c\ $ such that $\ \binom cn\ $ has at least $\ n\ $ different prime divisors $\ p\gt n$.


Of course it would be nice (but not realistic?) to compute these two sequences or give a sharp estimate. Even not so sharp bounds but with explicit scope of application like *for every* $\ n\ge 720\ $ *(or whatever the constant)* rather than *for almost all...* would be still interesting, I'd think, and possibly can find some applications (say, combinatorial). 


The properties under the consideration can be monotonicity of the sequences $\ \beta\ $ and $\ \gamma\ $ (including the deviations from the monotonicity); also the frequency of occurrences of prime numbers as $\ \beta(n)\ $ or $\ \gamma(n)\ $ respectively. Would they be more common in $\ \gamma\ $ then in $\ \beta\,$?



Prime divisors of the respectively minimal binomial coefficients
I'll just consider the second of the two problems posed. Changing notation slightly, it asks for the least natural number $n=n(k)$ such that $(n+1)(n+2)\cdots (n+k)$ is divisible by $k$ primes all larger than $k$. First I claim that $n(k) \le Ck^{e}$ for some constant $C$. To see this note that 
 $$ 
 \sum\_{n\le N} \sum\_{p>k, p|(n+1)\cdots (n+k)} 1 = \sum\_{k<p\le N} \sum\_{j=1}^{k} \sum\_{n\le N, p|(n+j)} 1 = 
 \sum\_{k< p \le N} k \Big( \frac{N}{p} +O(1)\Big), 
 $$ 
 and using $\sum\_{p\le x} 1/p = \log \log x + B + O(1/\log x)$ this is 
 $$
 = kN \log \frac{\log N}{\log k} + O\Big(\frac{N}{\log k}\Big).
 $$ 
 If $N >C k^e$ for some large $C$ then this exceeds $kN$, and we deduce that there is some $n \le N$ with $(n+1)\cdots (n+k)$ containing $k$ primes all larger than $k$. This easy argument may be found in Theorem 3.3 (part 3) of [Erdos and Turk](https://www.renyi.hu/~p_erdos/1984-17.pdf). 


My guess would be that $n(k) \ge k^{e-\epsilon}$ for any $\epsilon >0$, but this seems difficult to show. One can see easily that $n(k) \ge c k^2$ for some small positive constant $c$. To prove this, note that if $n\le k^2-k$ then each $n+j$ (with $1\le j\le k$) can have at most one prime factor larger than $k$. So if we can show that there are integers in $[n,n+k]$ that are $k$-smooth (all prime factors below $k$) then clearly $(n+1)\cdots (n+k)$ cannot be divisible by $k$ primes larger $k$. Choose $r=\lceil \sqrt{n+k}\rceil$ and $s = \lceil \sqrt{r^2 - (n+k)}\rceil$. Then $r^2-s^2$ can be checked to be in $[n,n+k]$ (if $n\le ck^4$ for a suitable $c$), and $r^2-s^2=(r+s)(r-s)$ is clearly $(r+s)$-smooth. This shows that if $n\le ck^2$ then the interval $[n,n+k]$ will contain $k$-smooth numbers. (In fact, it shows that the interval $[x,x+cx^{1/4}]$ contains $\sqrt{x}$-smooth numbers.) Erdos and Turk (see Theorem 3.3 part 1) showed that $n(k) \ge k^{2+\delta}$ for some small $\delta>0$, and this remains the best known. 


To show why the problem of getting lower bounds for $n(k)$ is difficult, let me isolate a consequence 
 which is not yet known. If the bound $n(k) \ge k^A$ holds for all large $k$, then every interval 
 $[x,x+x^{1/A}]$ must contain a $x^{1/A}$-smooth number. This consequence is only known for $A<2.5$; this follows from the argument involving $r^2-s^2$ given above, and was noted in a paper of [Friedlander and Lagarias](http://www.math.lsa.umich.edu/~lagarias/doc/short.pdf) (which appeared in J Number Theory). So proving the expected lower bound of $n(k) \ge k^{e-\epsilon}$ would have interesting, as yet unknown, consequences. 


