Let’s have $N_1=p^a$ where $p$ a prime number and $a$ a natural number. If we want to find all numbers from $1-N_1$ which are not divisible by $p$ we have to subtract all multiples of $p$ from $N_1$. The multiples are $1p,2p,...p^{a-1}p$. From this we conclude that the numbers not divisible by $p$ are $p^a-p^{a-1}$ or $N_1-N_1/p$. Let’s have the product of two primes $p^a q^b=N_2$ where $b$ a natural number. Now we have to find which numbers from $1-N_2$ are not divisible by $p,q$. In order to do so we have to subtract from $N_2$ the multiples $N_2/p$ and $N_2/q$. Now we have $(N_2-N_2/p)-N_2/q$, but some numbers are multiples of $pq$. We can subtract these multiples from $N_2/p$ or $N_2/q$. Let’s subtract them from $N_2/q$, giving $N_2/q-N_2/pq$. The last result has to be subtracted from the first term $(N_2-N_2/p)$. So we have $(N_2-N_2/p)-(N_2/q-N_2/pq)=F(N_2)$. So the number $F(N_2)$ is counting the numbers from $1-N_2$ which are not divisible by $p,q,pq$.
We obtain the second term by applying the ABROZ technique. This works as follows: we multiply the first term by $1/q$ and we obtain$ (N_2-N_2/p) 1/q=(N_2/q-N_2/pq)$. Let’s have $N_3=p^a q^b f^c$ where $ f$ a prime number and $c$ a natural number. We know how to get the first two terms and now have to obtain the third term which contains the multiples of $f$.
We again apply the technique, multiplying the two terms by $1/f$. So we have $(N_3-N_3/p)-(N_3/q-N_3/pq) 1/f=N_3/f-N_3/pf-N_3/qf-N_3/pqf$. To find $F(N_3)$ we subtract from the two terms the third term.
So we have: $F(N_3)=(N_3-N_3/p)-(N_3/q-N_3/pq)-(N_3/f-N_3/pf-N_3/qf-N_3/pqf)$.
We can continue this process indefinitely. The last equation can also be written as $F(N_3)=N_3[1-1/p-1/q+1/pq-1/f+1/pf+1/qf-1/pqf]$. The result inside the bracket can be obtained by multiplication of the three primes as follows: $(1-1/p)(1-1/q)(1-1/f)=[1-1/p-1/q+1/pq-1/f+1/pf+1/qf-1/pqf]$.
So we have $F(N_3)=N_3(1-1/p)(1-1/q)(1-1/f)$. We can make keys when$ F(N)/N$ is minimum or maximum.
My first question is: when is it easier to attack RSA, when $F(N)/N$ is minimum or maximum? Secondly, in what other areas of mathematics can we apply the maximum and minimum ratios?