I've been trying to solve this problem:
Find Euler's totient function of binomial coefficient C(n, m) = n! / (m! (n - m)!) modulo 10^9 + 7, m <= n < 2 * 10^5.
One of my ideas was that first, we can precalculate the values of phi(i) for all i from 1 to n in linear time, also we can calculate all inverses to numbers from 1 to n modulo 10^9 + 7 using, for example, Fermat's little theorem. After that, we know, that, in general, phi(m * n) = phi(m) * phi(n) * (d / fi(d)), d = gcd(m, n). Because we know that gcd((x - 1)!, x) = 1, if x is prime, 2 if x = 4, and x in all other cases, we can calculate phi(x!) modulo 10^9 + 7 in linear time. However, in the last step, we need to calculate phi(n! / ((m! (n - m)!), (if we already know the function for factorials), so, if we are using this method, we have to know gcd(C(n, m), m! (n - m)!), and I don't know how to find it.
I've also been thinking about factorizing the binomial coefficient, but there seems no efficient way to do this.
Any help would be appreciated.
First, factorize all numbers 1..(2*10^5) as products of prime powers.
Now, factorize n!/k! = n(n-1)(n-2)...(n-k+1) as a product of prime powers by multiplying together the factors of the individual parts. Factorize (n-k)! as a product of prime powers. Subtract the latter powers from the former (to account for the divide).
Now you've got C(n, k) as a product of prime powers. Use the formula phi(N) = N * prod(1 - 1/p for p|N) to calculate phi(C(n, k)), which is straightforward given that you've computed the a list of all the prime powers that divide C(n, k) in the second step.
For example:
phi(C(9, 4)) = 9*8*7*6*5 / 5*4*3*2*1
9*8*7*6*5 = 3*3 * 2*2*2 * 7 * 3*2 * 5 = 7*5*3^3*2^4
5*4*3*2*1 = 5 * 2*2 * 3 * 2 * 1 = 5*3*2^3
9*8*7*6*5/(5*4*3*2*1) = 7*3^2*2
phi(C(9, 4)) = 7*3^2*2 * (1 - 1/7) * (1 - 1/3) * (1 - 1/2) = 36
I've done it in integers rather than integers mod M, but it seems like you already know how division works in the modulo ring.