Euler phi function and powers of two For small values of  n$\ge$1  it appears that one has the inequality
$\phi(2^n)$ $\le$ $\phi(2^n + 1)$.
However, it seems unlikely that this would hold for all n .
Question:  Are there any explicit values of n known for 
              which the inequality is violated ?
Thanks

For each prime factor $p|2^n+1$ you get a reduction of $\frac {p-1}P$ in the $\varphi$ function. The $p$ you are looking for are ones for which $2^n\equiv -1 \bmod p$. Then for odd $k$, $2^{kn}\equiv -1 \bmod p$. So if we work with odd exponents we can collect primes - the question is can we collect enough small primes.
Now we have $2^1\equiv -1\bmod 3, 2^5\equiv -1 \bmod 11, 2^7\equiv -1 \bmod 43$ and the list of powers and primes continues $(1:3), (5:11), (7:43), (9:19), (11:683), (13:2731), (15:331)$ - for these the relevant factor is approximately $0.5581$ and we need it below $0.5$
No doubt someone can search more efficiently than I can to provide a definitive conclusion.
