Let $\alpha > 0$ be a real number and let us consider the set $S(\alpha)$ of those natural numbers $n$ such that the fractional part of $\alpha \cdot n$ "begins" with the representation of $n$ (in base $10$). Formally,
$$
S(\alpha) = \{k\in \mathbb{N}:k=\lfloor \operatorname{frac}(\alpha k)\cdot 10^{1+\lfloor\log_{10} k \rfloor}\rfloor\}
$$
where $\operatorname{frac}(x) = x-\lfloor x\rfloor$, for $x>0$, denotes the fractional part of $x$.
For example, $57211\in S(\sqrt{2})$, since $57211\sqrt{2} = 80908.\underline{57211}692\cdots$.
If $\alpha$ is an irrational number, we know that $\operatorname{frac}(\alpha\cdot k)$ is uniformly distributed in $(0,1)$, so, using a rough heuristic argument based on the fact that $\sum\frac{1}{k}$ diverges, we may expect $S(\alpha)$ to have sparse but infinite elements.
A few computations relative to well-known irrational constants support this intuition. For example, we have
$$
S(\pi) = \{ 1,2,38,76,946,24996,3595182,61864425177,\dots\}\,,
$$
and
$$
S(\sqrt{2})=\{ 772,9792,57211,535090,6101272,65645433,9169209625,16835518309,\dots\}\,,
$$
but what really baffles me is
$$
S(e)=\{ 5,191,\\ 1100,1210,1320,1430,1540,1650,1760,1870,1980,2090,2200,2310,2420,2530,2640,
2750,\\ 2860,2970,3080,3190,3300,3410,3520,3630,3740,3850,3960,4070,4180,4290,4400,4510,\\
4620,4730,4840,4950,5060,5170,5280,5390,5500,5610,5720,5830,5940,6050,6160,6270,\\ 6380,
6490,6600,6710,6820,6930,7040,7150,7260,7370,7480,7590,7700,7810,7920,8030,\\ 8140,8250,
8360,8470,8580,8690,8800,8910,9020,9130,9240,9350,9460,9570,9680,\\ 1865037,5422244075, \dots\}\,.
$$
As you can see there are $79$ terms between $10^3$ and $10^4$, all divisible by 10.
My question is: is the behavior of $S(e)$ just a coincidence or the nature of $e$ has something to do with it?
Best Answer
Note that in order for a $k\in S(e)$, $k\cdot e$ must have digits of $k$ in its decimals. For example, $4180e=11362.\underline{4180}4...$
Also observe that there is a $110$ gap between consecutive $k$. It turns out $110e=299.01100113...\approx 299+110*10^{-4}$ and its digits after the $1100$ is small.
Because it just so happens that $1100\in S(e)$
Then $(1100+110m)e\approx n+(1100+110m)*10^{-4}$, where $n$ is the integral part of the number.
it follows that
$$(1100+110m)e -\lfloor (1100+110m)e\rfloor \approx 0.(1100+110m)$$
Due to the fact that $110e$ has smaller value ($113$) after the $1100$ sequence in its digit. (And $1100$ being in $S(e)$, which is kind of a coincidence)
Edit: what I meant by the digits after $1100$ are small is that they won't affect the higher digits until about $100$ multiple of $110e$. For example, $9790$ doesn't work because $89*0.00000113\approx 0.001$, causing a change in digit in the fourth place