Does there exist any formula for finding out $\phi(n!)$ I don't Know if it the right place to ask a question of this type here:
Does there exist any formula for finding out $\phi(n!)$ where $\phi(n)$ is the Euler's phi function.
 A: $\varphi(n!)=n!\prod\limits_{p\leq n}\frac{p-1}{p}$.
In general $\varphi(n)$ is $n$ multiplied by the product $\frac{p-1}{p}$ over all the primes dividing $n$, in this case these primes are the primes less than or equal to $n$.
A: At least two formulas are known, the first of which will be mentioned in plenty other answers besides this one.
Then there is a formula of a kind called a "recurrence relation," the most famous of these perhaps being the formula for the Fibonacci numbers which uses the previous two Fibonacci numbers to obtain the next Fibonacci number. For this formula for $\phi(n!)$, we need to know $\phi((n - 1)!)$ and whether $n$ is prime or composite. If $n$ is prime, then $\phi(n!) = \phi((n - 1)!)(n - 1)$. If $n$ is composite, $\phi(n!) = \phi((n - 1)!)n$.


*

*Obviously $\phi(1!) = \phi(2!) = 1$.

*Then 3 is prime, so $\phi(3!) = 2\phi(2!) = 2$.

*4 is composite, so $\phi(4!) = 4\phi(3!) = 8$.

*5 is prime, so $\phi(5!) = 4\phi(4!) = 32$.

*6 is composite, so $\phi(6!) = 6\phi(5!) = 6 \times 32 = 192$.

*7 is prime, so $\phi(7!) = 6\phi(6!) = 6 \times 192 = 1152$.
I wish I could take credit for this, but I just took it from Sloane's A048855. Well, first I had Mathematica calculate a few terms for me, using Table[EulerPhi[n!], {n, 10}], then I copied and pasted the output into the OEIS search box.
