Number that is equal to the cesaro mean of its digits Consider a number, $x > 0$, of the form $x = b_1.b_2b_3...$, where $b_i$ is the $i$th digit.
Let
$$
 C(x)= \lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^{N}b_n
$$
be the Cesaro mean of the digits of $x$. For example, for $x=3.33333...$, $C(x) = 3$, and for $3.4444...$, $C(x) = 4$.
Does there exist an $x$ that is equal to the Cesaro mean, $C(x)$ of its digits? That is,
$$
x= C(x)?
$$
It can be seen that $x \leq 9$. I am not sure if any fixed point theorems can be applied here to construct one such number.
 A: I'm pretty sure the following greedy construction works, although it would be great if others could contribute to a proof.
Define a sequence $\{d_i\}$ of decimal digits and the associated terminating rational numbers $r_j = \sum_{i=1}^j \frac{d_i}{10^i}$ recursively as follows: set $d_1=1$, and given $d_1,\dots,d_k$, set $d_{k+1}=1$ if $\frac 1k(d_1+\cdots+d_k) \le r_k$ and $d_{k+1}=0$ otherwise. Then $d_2=\cdots=d_{10}=0$, and $d_{11}=1$ since we used $\le$ instead of $<$ in the definition of $d_{k+1}$, then $d_{12}=\cdots=d_{20}=0$ and $d_{21}=1$, and indeed $d_i=1$ for $i=11,21,31,41,\dots$ and this pattern repeats until $d_{10^{10}}=1$ instead of $0$....
Anyway, it seems that $x=\lim\limits_{j\to \infty} r_j$ should have the property that it equals the Cesàro mean of its digits.
Assuming this is right, the construction should be pretty robust to changes: it should work for any two digits $0\le m<n\le 9$ in place of $0$ and $1$, as long as we give $x$ an integer part that is between $m$ and $n-1$ inclusive; and it should work with any finite number of the $d_i$ prescribed arbitrarily as well. In particular, this generalization would show that the set of real numbers that equal the Cesàro mean of their digits is dense in $[0,9]$. One could probably even show that this set is uncountable with another generalization step—say we could choose to set either $d_{k+1}$ or $d_{k+2}$ to be the larger digit, and we had this binary choice each time the larger digit was about to appear.
A: Here's a proof that no rational number equals the Cesàro mean of its digits.
Let $r = \frac ab$ in lowest terms, and write $b=cd$ where $c=2^j5^k$ for some $j,k\ge0$ and $\gcd(d,10)=1$. We can assume that $d>1$, or else the decimal expansion of $r$ terminates and the Cesàro mean of its digits equals $0$.
Then the period of the repeating part of the decimal expansion of $r$ equals the multiplicative order of $10$ modulo $d$. In particular, the period must be a divisor of $\phi(d)$, and therefore $\phi(d)$ times the Cesàro mean must be an integer.
However, $\phi(d)r = \frac ac \frac{\phi(d)}d$ cannot be an integer: if $p$ is the largest prime dividing $d$, then after reducing $\frac{\phi(d)}d$ to lowest terms, its denominator is divisible by $p$, and this cannot be canceled by the numerator of $\frac ac$, since $\frac ab$ being in lowest terms implies that $\gcd(a,d)=1$.
