Numerical analysis of the first several hundred n suggests the following inequality:
$\varphi(3^n2) \ge 2\cdot3^{n1}$
Numerical analysis of the first several hundred n suggests the following inequality: $\varphi(3^n2) \ge 2\cdot3^{n1}$ 


Take $n=382315009082231724951830011$. Then $3^n2$ is divisible by the primes $5, 19, 23, 47, 71, 97, 149, 167, 173, 263, 359, 383, 389, 461, 479, 503, 557$. Furthermore, $\varphi(3^n2)<2/3\cdot(3^n2)<2\cdot 3^{n1}$. The following Sage code verifies this examples. I believe that this is close to a minimal counterexample. sage: n=382315009082231724951830011 sage: l=[5, 19, 23, 47, 71, 97, 149, 167, 173, 263, 359, 383, 389, 461, 479, 503, 557] sage: set(3.powermod(n,p) for p in l) set([2]) sage: prod(11/p for p in l).n() 0.666250824539016 As there are some speculations about how to find such an example, here is the (not really clever) Sage code which greedily collects the congruences for $n$ which do not contradict each other: p,n,Q = 5,3,4 lp = [p] s = 11/p while True: p=p.next_prime() e=IntegerModRing(p)(3).multiplicative_order() l=[z for z in [1..e1] if 3.powermod(z,p) == 2] if len(l) == 0: continue b=l[0] if (bn) % e.gcd(Q) != 0: continue QQ=Q.lcm(e) n=CRT_list([n,b],[Q,e]) % QQ Q=QQ s*=(11/p) lp.append(p) print p,s.n() if 2/3>s: print n,lp break 

