I know that Euler totient function is multiplicative, that is $\phi(mn) = \phi(m) * \phi(n)$. And obviously it would't have additive feature.
So i'm looking for any existing formula which facilitate calculation of $\phi(N)$ if $\phi(N-1)$ is known. Is there any?
Updated
On this page, I found an interesting conjecture, which is hold for some numbers:
For all $n=>4$ it always exists an integer $k, 1<=k<=n-1$ such that:
$\phi(n^2-k^2)=(n-1)^2-k^2$
Solving that would lead to $\phi((n-k)(n+k))=(n-1)^2-k^2$ $\phi(n-k)*\phi(n+k)=(n-1)^2-k^2$ $\phi(n+k)=\frac{(n-1)^2-k^2}{\phi(n-k)}$
and if $n-k$ is prime, then $\phi(n-k) = n-k-1$ $\phi(n+k)=\frac{(n-1)^2-k^2}{n-k-1}=\frac{(n-1-k)(n-1+k)}{n-k-1}$ $\phi(n+k)=n+k-1$
and since $\phi(n+k)=n+k-1$ then $n+k$ should be prime.
Examples of $n, k$ for which the above conjecture holds:
$n = 27, k=4$
$n = 300, k=31$
$n= 1500, k=1231$
Finding an order among solutions, would lead to a rewritten conjecture:
If $n-k$ is prime then $n+k$ should be prime.