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
 A: Mark Bennet inspired me, so thanks in advance and a (+1) as a tribute.
We may collect a large number of primes of the form $8k+3$, say $p_1,\ldots,p_N$.
$\frac{p_k-1}{2}$ is odd and by quadratic reciprocity, $\left(\frac{2}{p_k}\right)=-1$, so if we take:
$$ n=\prod_{k=1}^{N}\frac{p_k-1}{2}\tag{1} $$
we have:
$$ 2^n+1 \equiv \left(2^{\frac{p_k-1}{2}}\right)^d+1 \equiv (-1)^d+1\equiv 0\pmod{p_k}\tag{2}$$
and every $p_k$ divides $2^n+1$. Since
$$ \sum_{p\equiv 3\pmod{8}}\frac{1}{p}\tag{3} $$
is a divergent series,
$$ \prod_{p\equiv 3\pmod{8}}\left(1-\frac{1}{p}\right) = 0,\tag{4} $$
so the inequality fails for infinitely many $n$s. 
Thanks to Thomas Andrews, we also have an explicit counter-example:
$$\prod_{\substack{p\equiv 3\!\!\pmod{\!\!8}\\p\leq 467}}\!\!\frac{p-1}{2}=118,036,297,108,708,732,142,419,220,832,488,800,078,125.$$
Mark Bennet gives a way smaller counter-example:
$$ n=3^4⋅5^2⋅7⋅11⋅13⋅17⋅23⋅29⋅41⋅53=49,945,180,460,175.$$
A: 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.
