Totient function gives us the count of of all numbers(k) upto n which satisfy the equation gcd(k,n)=1. How can I find the count of all such k's which satisfy gcd(k,n)=d for some natural number d?
2026-03-25 06:17:35.1774419455
Generalization of Totient Function
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Assuming that $d$ divides $n$, then $\gcd(k,n)=d \implies \gcd(k/d, n/d) = 1$ so the count of such values you'd be looking for is $\phi(n/d)$.