Simplified totient for primorials I have found a function that computes Euler's totient function exactly for primorial inputs as follows:
$$\phi(P_n \#) = \left(\prod_{a=1}^n P_a - 1\right)$$
I am able to prove that this function works for every possible primorial input; the problem is that it would be incredibly presumptuous to think that this hasn't been found by someone already without first doing an exhaustive literature search.
Has anyone heard of this function before, and where would I go to read more about it?
 A: The totient function is known to be multiplicative, which means that $\phi(pq) = \phi(p)\phi(q)$ whenever $p$ and $q$ are "relatively prime"—that is, whenever $p$ and $q$ have no common factor greater than 1.  This obviously occurs whenever $p$ and $q$ are actually prime, so in fact
$$\begin{align}\phi(p_1p_2\cdots p_i)  &= \phi(p_1)\phi(p_2)\cdots\phi(p_i)\\
& = (p_1-1)(p_2-1)\cdots(p_i-1)\end{align}$$
for any distinct primes $p_1, p_2, \ldots p_i$.  This covers your formula, since a primorial is a product of distinct primes.
A generalization,
$$\phi\bigl(p_1^{n_1}p_2^{n_2}\cdots p_i^{n_i}\bigr) = \phi\bigl(p_1^{n_1}\bigr)\phi\bigl(p_2^{n_2}\bigr)\cdots \phi\bigl(p_i^{n_i}\bigr)$$
follows immediately.
Since $\phi\bigl(p^n\bigr)$ is easy to calculate—it is just $(p-1)p^{n-1}$—one can easily calculate $\phi()$ whenever the prime factorization of the argument is known, as it is in your case. For example, what is $\phi(1000)$?  This is $\phi\bigl(2^3 5^3\bigr) = \phi\bigl(2^3\bigr)\phi\bigl(5^3\bigr) = 1\cdot2^2\cdot4\cdot5^2 = 400$.
This was all known to Euler; sorry.
You asked "where would I go to read more about it?" I suggest that if your're interested that you try looking into a few elementary number theory textbooks until you find one you like; any book that discusses $\phi$ at all will surely mention all of this quite early on. In particular it appears in Concrete Mathematics, by Graham, Knuth, and Patashnik, and although that is not a number theory textbook you might enjoy it.
[ It's much too late, but I wish I had added that I think you should be proud of your discovery, which is not only correct, but legitimate and serious mathematics. ]
