The maximal size of between $\varphi(n)$ divided by $\lambda(n)$. I want to find $$f(n) = \max\left\{\frac{\varphi(k)}{\lambda(k)} : 1 \leq k \leq n\right\}$$
In other words, I want to find the maximal value of $\frac{\varphi(k)}{\lambda(k)}$ when $k$ is restricted.
$\lambda(n)$ is the Carmichael function, the smallest $k$ such that for all $a$ such that $\gcd(a,n)=1$ the following congruence holds: $a^{k} \equiv 1 \mod n$. $\varphi(n)$ is the Euler-phi function. Note that $\lambda(n) \mid \varphi(n)$. 
Since I don't expect $f(n)$ to have a nice expression in terms of elementary and number theoretic functions, I started to search for some lower bounds. I found that $$\frac{\varphi(2^k \cdot p_{k+1}\#)}{\lambda(2^k \cdot p_{k+1}\#)} \geq 2^k$$
Here $k\#$ is the primorial function. 
Therefore I know that the fraction is unbounded, and that $$f(n) \in \Omega(n^{1/ \ln \ln n})$$
However $n^{1/ \ln \ln n}$ is still bounded above by all power functions with a positive power. Does anyone have an asymptotically better bound or a proof that this is the best bound?
I looked to OEIS, which has this sequence A034380. I found that the smallest $n$ such that $f(n)\geq 108$ is $n=3591$, the smallest $n$ such that $f(n)\geq 128$ is $n=5440$. However my method gives $1241560320$ as a number such that it is bigger than 128. Clearly, there is some room for improvement. 
Conjecture: Will Jagy found strong numerical evidence that $f(n)$ eventually outgrows $n^{1-\varepsilon}$ for every $\varepsilon > 0$. A proof (or disproof) of this would be very welcome.
 A: Got a reference on MO; see https://mathoverflow.net/questions/210144/number-of-primes-one-larger-than-divisors-of-a-fixed-number-which-is-lcm-of-1-2#comment520981_210144   and 
http://www.math.drexel.edu/~eschmutz/PAPERS/lambda.pdf
In Theorem $1$ on the first page of the Erdos-Pomerance-Schmutz article, they announce the existence of a constant $c$ and a sequence of numbers $n$ going to $\infty,$ such that
$$ \lambda(n) < \left( \log n \right)^{c \log \log \log n}.  $$
On the other hand, for all $n \geq 3, $ we find in Rosser and Schoenfeld (1962) that
$$ \varphi(n) > \frac{n}{ e^\gamma \log \log n + \frac{3}{\log \log n}}. $$
So, on the special sequence in Erdos-Pomerance-Schmutz, we find
$$\frac{ \varphi(n)}{\lambda(n)} > \frac{n}{ \left( e^\gamma \log \log n + \frac{3}{\log \log n} \right) \left(   \left( \log n \right)^{c \log \log \log n} \right)  } $$
which is eventually bigger than any $n^{1-\delta}.$ I am not at all clear what their sequence is. 
A: This is conditional on Dickson's conjecture, but may be of interest.
You may choose any
admissible prime k-tuple $(b_1=0,b_2,b_3,\ldots,b_k)$. By definition the $b_i$ avoid some residue modulo every prime, and hence so do $\{(2+b_i)n+1\}$. Then it is a consequence of Dickson's conjecture that $q_i=(2+b_i)n+1$ are simultaneously prime for $1\le i \le k$ for infinitely many choices of $n$.
Write $N=\prod q_i$, then
$$
\begin{align}
N &= \prod_{i=1}^k\left((2+b_i)n+1\right) \\
& < (n+1)^k\prod_{i=1}^k (2+b_i) \\
&= C_1 (n+1)^k
\end{align}
$$
and
$$
\begin{align}
\varphi(N)&=\prod_{i=1}^k (q_i-1) \\
&= n^k \prod_{i=1}^k (2+b_i) \\
&= C_1 n^k
\end{align}
$$
and
$$
\begin{align}
\lambda(N) &= \operatorname{lcm} \left(2n,(2+b_2)n,\ldots,(2+b_k)n\right) \\
& \le n\operatorname{lcm}\left(2+b_2,2+b_3,\ldots,2+b_k\right) \\
&= C_2 n
\end{align}
$$
Where $C_1$ and $C_2$ depend on the choice of k-tuple but not on $n$.
Thus
$$
f(N) = \frac{\varphi(N)}{\lambda(N)} \ge \frac{C_1}{C_2}n^{k-1} = \Theta\left(N^{1-1/k}\right)
$$
and by appropriate choice of $k$ can be made to grow faster than $N^{1-\epsilon}$ for any $\epsilon>0$.
For example, $(0,2,6,12,14,20,24,26,30,32)$ is an admissible 10-tuple of prime differences.
We find simultaneous primes $q_i$ when
$$
n = 190103265 \\
N \approx 1.540\times10^{94} \\
\varphi(N) \approx 1.540\times 10^{94} \\
\lambda(N) = 544544n \approx 1.035\times10^{14}\\
f(N) \approx 1.488\times10^{80} \\
\frac{\log f(N)}{\log N} = 0.8512\ldots
$$
and when
$$
n = 5092580565 \\
N \approx 2.93\times10^{108} \\
\varphi(N) \approx 2.93\times 10^{108} \\
\lambda(N) = 544544n \approx 2.77\times10^{15}\\
f(N) \approx 1.057\times10^{93} \\
\frac{\log f(N)}{\log N} = 0.8576\ldots
$$
For larger $n$ that yield solutions we will have $\log_N f \to 9/10$.
Similarly, for a larger choice of $k$ we should be able to make $\log_N f \to 1-1/k$, though in practice it is challenging to find examples.
