# How to calculate “gcd product” $\operatorname{gcdp}(n,m)=\gcd(n,1)\gcd(n,2)\cdots\gcd(n,m)$

Given two numbers $m$ and $n$ how can we calculate the gcd product of any two numbers i.e, $\operatorname{gcd p}(n,m)=\gcd(n,1)\gcd(n,2)\cdots\gcd(n,m)$ where gcd is the greatest common divisor?

Can this be solved using Euler Totient function?

My Approach:-

For any prime number $n$, $\operatorname{gcdp}(n,m)=n^{\operatorname{floor}(m/n)}$.

Is there any way to solve it for any non-prime $n$?

Let $$f_m(n)=\gcd(n,1)\gcd(n,2)\gcd(n,3)\cdots \gcd(n,m).$$ To extend your computation of $f_m(n)$ beyond the primes, note that fox fixed $n$ the function $f_m$ is multiplicative: If $\gcd(a,b)=1$, then $f_m(ab)=f_m(a)f_m(b)$.
Thus it is enough to know $f$ at prime powers. In particular, since you know $f_m(p)$ for prime $p$, you know $f_m(n)$ for all square-free $n$.
Now let us get a handle on $f_m(p^2)$, where $p$ is prime. We will pick up a $p$ for every number $i\le m$ such that $i$ is divisible by $p$. We will pick up an extra $p$ for every number $i\le m$ which is divisible by $p^2$. Thus the total number of $p$'s that we pick up is $\left\lfloor\frac{m}{p}\right\rfloor+\left\lfloor\frac{m}{p^2}\right\rfloor$. That gives us the exponent of $p$ in $f_m(p^2)$.
Exactly the same reasoning can be used to find the exponent of $p$ in $f_m(p^3)$. We get $\left\lfloor\frac{m}{p}\right\rfloor+\left\lfloor\frac{m}{p^2}\right\rfloor+ \left\lfloor\frac{m}{p^3}\right\rfloor$. The idea readily generalizes to $f_m(p^k)$.
Thus we can get an explicit expression for $f_m(n)$, once we know the prime power factorization of $n$.