An infinite sum based on the mod-parity of Euler's totient function Let $\bmod( m,k )$ be the remainder when $m$ is divided by $k$: $0,1,\ldots,m{-}1$.
Let $\phi(n)$ be Euler's totient function: the number of relatively prime numbers smaller than $n$.
So for
$$n=(1,2,3,4,5,6,7,8,9,10) \;,$$
$$
\phi(n)=(1,1,2,2,4,2,6,4,6,4) \;.
$$
Define 
$$
c_3 = \sum_{n=1}^\infty \frac{\bmod( \phi(n),3)}{3^n} \;.
$$
For $n=1,\ldots,10$,
$$
\bmod( \phi(n),3)=(1,1,2,2,1,2,0,1,0,1) \;,
$$
and the sum up to $n{=}10$ is $\frac{32491}{59049} \approx 0.5502379380$.
Because $\sum 2/3^n = 1$, $c_3 < 1$.

Q. Has this sum $c_3$ been studied? Is it irrational?

Note that $c_2 = \frac{3}{4}$ (because $\phi(n)$ is even for $n>3$),
but $c_k$, $k > 2$ seems unclear. So I ask for information for the first
interesting sum, when $k{=}3$.
 A: $c_k$ is provably irrational for all $k>2$.
Note that for any fixed $k>2$, there are infinitely many $n$ with $k \nmid \phi(n)$: if $k$ has an odd divisor $q$ then pick $n$ to be a prime $\equiv 2 \pmod q$; otherwise $4 \mid k$, in which case pick $n$ to be a prime $\equiv 3 \pmod 4$.
It's easy to see that also $k \mid \phi(n)$ for infinitely many $n$; in particular, the base-$k$ digits of $c_k$ correspond as expected to the mod-$k$ residues of $\phi(n)$ (there aren't degeneracies like 0.12323499999... = 0.123235 to fuss over).
Suppose that $c_k$ is rational.  Then its base-$k$ digits are eventually periodic with some period $M$.  Since there are infinitely many $n$ with $k \nmid \phi(n)$, $c_k$ has at least one non-zero digit within this period.  In particular, there is some $i>0$ such that $k \nmid \phi(i+Mj)$ for all $j\ge 0$ (an arithmetic progression of non-zero digits).
But this is impossible: let $p$ be any prime congruent to $1 \pmod k$ which is larger than $M$ (and thus relatively prime to $M$).  Then $\{Mj: j \in \mathbb N\}$ forms a complete residue set mod $p$, which means $p \mid (i+Mj_0)$ for some $0 \le j_0 < p$. But then $\phi(i+Mj_0) \equiv 0 \pmod k$, a contradiction.
I think one could probably show more strongly that the number of $n$ for which $k \nmid \phi(n)$ is sublinear, and this'd give another proof of irrationality (the non-zero digits are too sparse).
