I read this paper by P. Erdos, page 2. I didn't understand it. How do I determine the number divisors of $n!$ ? I'd like an example application, for example if I want to determine the number divisors of $150!$ then what can I do?
Thank you for any help.
On
If you want to apply some of his ideas to a specific $n$, then I would suggest starting with the formula at the beginning of section 2 that gives the expression for $n!$ as a product of $p^{w_p(n)}$ terms (the prime factorisation). The formula for $w_p(n)$ is simply expressing the fact that every $p-th$ number ranging from $1,...,n$ has a factor of $p$ and every $p-th$ one of these has an additional factor of $p$, etc. (or count multiples of $p,p^2,p^3,...$ and apply GPIE since every multiple of $p^2$ is also a multiple of $p$, and so on).
Then you get an example such as:
$20! = 2^{10+5+2+1}\cdot3^{6+2}\cdot5^{4}\cdot7^{2}\cdot11\cdot13\cdot17\cdot19 = 2^{18}\cdot3^{8}\cdot5^{4}\cdot7^{2}\cdot11\cdot13\cdot17\cdot19$
Once the prime factorisation has been determined, then $d(n!)$ is a product of the $(1+w_p(n))$ terms, since each divisor is essentially an independent choice of each of the prime exponents in the factorisation. Erdös has just taken the $log$ of this relation to get a summation instead of a product.
Then, for example
$d(20!) = 19 \cdot 9 \cdot 5 \cdot 3 \cdot 2^4 = 41040$
Suppose $m=p_1^{e_1}\cdots p_k^{e_k}$ where $p_1,\cdots,p_k$ are distinct primes and $e_1,\cdots,e_k>0$. Then any divisor of $m$ will look like $p_1^{r_1}\cdots p^{r_1}$ with $0\le r_i\le e_i$ for each $i=1,\cdots,k$. With $i=1$, there are a total of $e_1+1$ choices for $r_1$, with $i=2$ there are $e_2+1$ choices for $r_2$, and so on. Therefore the total number of divisors equals $(e_1+1)(e_2+1)\cdots(e_k+1)$.
Of course the only prime factors of $n!$ will be primes $p\le n$. We seek to find the exponent of $p$ in the prime factorization of $n!$. An argument over here at cut-the-knot shows it is
$$v_p(n!)=\left\lfloor\frac{n}{p}\right\rfloor+\left\lfloor\frac{n}{p^2}\right\rfloor+\left\lfloor\frac{n}{p^3}\right\rfloor+\cdots.$$
Therefore,
$$\sigma_0(n!)=\prod_{p\le n}\left(1+\sum_{k\ge1}\left\lfloor\frac{n}{p^k}\right\rfloor\right).$$
Let's try to compute this with $n=150$. Typing in
primes less than 150into Wolfram Alpha, I get that the relevant primes are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149}.Prime $p=2$. Since $128\le 150<256$ we have $$v_2(150!)=\left\lfloor\frac{150}{2}\right\rfloor+\left\lfloor\frac{150}{4}\right\rfloor+\left\lfloor\frac{150}{8}\right\rfloor+\left\lfloor\frac{150}{16}\right\rfloor+\left\lfloor\frac{150}{32}\right\rfloor+\left\lfloor\frac{150}{64}\right\rfloor+\left\lfloor\frac{150}{128}\right\rfloor$$ $$=75+37+18+9+4+2+1=146$$
Prime $p=3$. Since $81\le150<243$ we have $$v_3(150!)=\left\lfloor\frac{150}{3}\right\rfloor+\left\lfloor\frac{150}{9}\right\rfloor+\left\lfloor\frac{150}{27}\right\rfloor+\left\lfloor\frac{150}{81}\right\rfloor$$ $$=50+16+5+1=72$$
Prime $p=5$. Since $125\le150<625$ we have $$v_5(150!)=\left\lfloor\frac{150}{5}\right\rfloor+\left\lfloor\frac{150}{25}\right\rfloor+\left\lfloor\frac{150}{125}\right\rfloor=30+6+1=37 $$
Prime $p=7$. Since $49\le150<343$ we have $$v_7(150!)=\left\lfloor\frac{150}{7}\right\rfloor+\left\lfloor\frac{150}{49}\right\rfloor=21+3=24 $$
Prime $p=11$. Since $121\le150<1331$ we have $$v_{11}(150!)=\left\lfloor\frac{150}{11}\right\rfloor+\left\lfloor\frac{150}{121}\right\rfloor=13+1=14 $$
For any prime $p\ge13$, we have $150<p^2$, and so $v_p(150!)=\lfloor150/p\rfloor$ for those primes.
Multiplying together, we compute
$$\begin{array}{l} (146+1)\times(72+1)\times(37+1)\times(24+1)\times(14+1) \\ \times(11+1)\times(8+1)\times(7+1)\times(6+1)\times(5+1) \\ \times(4+1)^2\times(3+1)^3\times(2+1)^6\times(1+1)^{14} \end{array}$$
$$=106~043~863~583~843~942~400~000.$$
(I checked this in Mathematica - it is correct.)