Q:

Question about $ m = \prod_i (p_i-1)^{b_i} $ and the unique factorization theorem.

The fundamental theorem of arithmetic says that every integer $n>1$ is of the form
$$ n = \prod_i {p_i}^{a_i} $$
where $p_i$ is the $i$ th prime and $a_i$ is a nonnegative integer.
My question is how many $m$ satisfy $ 1<m<n $ and 
$$ m = \prod_i {(p_i-1)}^{b_i} $$
where $p_i$ is the $i$ th ODD prime and $b_i$ is a nonnegative integer.
Lets call $f(n)$ the amount of $m$ that satisfy the above.
What is a good approximation to $f(n)$ ?

A:

You are asking for the counting function of the semigroup generated by the "shifted primes", the numbers $p-1$. Kevin Ford, 1998, showed that this counting function is on the order of $${x\over\log x}\exp(C_1(L_3(x)-L_4(x))^2+C_2L_3 (x)+C_3L_4(x))$$ where I write $L_3(x)$ and $L_4(x)$ for $\log\log\log x$ and $\log\log\log\log x$, respectively. See page 9 of this set of slides. I don't have bibliographic details for the 1998 paper. 
EDIT: There is a link to the paper on Kevin Ford's website, so anyone who wants to can now have a look at the proof. The result is given as Corollary 15. The bibliographic details are Kevin Ford, The distribution of totients, The Ramanujan Journal 2, no. 1-2 (1998) 67-151, but the link goes to "the updated 2012 version, with various corrections and simplifications to the original paper." 

