TITLE: Poisson like distribution in $\left(\sum_{k=0}^{n-1} x^{2^k}\right)^{n}$?
QUESTION [5 upvotes]: Given $y_{n}(x)=\left(\sum_{k=0}^{n-1} x^{2^k}\right)^{n}$. An example ($n=5$) may look like
$$
y_5(x)={x^{80}}+5 x^{72}+5 x^{68}+5 x^{66}+5 x^{65}+10 x^{64}+20 x^{60}+20 x^{58}+20 x^{57}+20 x^{56}+20 x^{54}+20 x^{53}+40 x^{52}+20 x^{51}+40 x^{50}+30 x^{49}+35 x^{48}+60 x^{46}+60 x^{45}+60 x^{44}+60 x^{43}+80 x^{42}+50 x^{41}+61 x^{40}+60 x^{39}+100 x^{38}+90 x^{37}+85 x^{36}+70 x^{35}+95 x^{34}+65 x^{33}+75 x^{32}+{120 x^{31}}+120 x^{30}+100 x^{29}+110 x^{28}+100 x^{27}+90 x^{26}+90 x^{25}+100 x^{24}+100 x^{23}+90 x^{22}+70 x^{21}+66 x^{20}+70 x^{19}+55 x^{18}+65 x^{17}+75 x^{16}+60 x^{15}+50 x^{14}+50 x^{13}+40 x^{12}+30 x^{11}+31 x^{10}+25 x^{9}+15 x^{8}+10 x^{7}+5 x^{6}+x^{5}
$$
If you plot exponents of addends of $y_n$ against prefactors, this looks (for $n=7$) like 
$\hskip1.7in$
Can this be described by a kind of Poisson distribution for general $n$?

REPLY [3 votes]: This is not a Poisson distribution, but is related. It is a rather interesting distribution, but let me not disclose it at the moment.
First note that, up to the multiple $n^n$, this is the probability generating function for the sum of $n$ iid random variables, which are uniformly distributed on the set $\{1,2,\dots,2^{n-1}\}$. 
Dividing by $2^{n-1}$, we get the sum of $n$ iid random variables $\xi_{1,n},\dots,\xi_{n,n}$, uniformly distributed on the set $\{1,2^{-1},2^{-2}\dots,2^{1-n}\}$. Then the characteristic function of the sum $S_n = \xi_{1,n}+\dots+\xi_{n,n}$ is
$$
\varphi_{S_n}(t) = \varphi_{\xi_{1,n}}(t)^n = \left(\frac1n \sum_{k=0}^{n-1}e^{it2^{-k}}\right)^n = \left(1+ \frac1n \sum_{k=0}^{n-1}\big(e^{it2^{-k}}-1\big)\right)^n\\
\to \exp\left\{\sum_{k=0}^{\infty}\big(e^{it2^{-k}}-1\big)\right\} = \prod_{k=0}^{\infty}\exp\left\{e^{it2^{-k}}-1\right\},\quad n\to\infty.
$$
Therefore, $$S_n\overset{w}{\longrightarrow} S = \sum_{k=0}^\infty \frac{\zeta_k}{2^k},\quad n\to\infty,$$
where $\zeta_k$ are iid with Poisson(1) distribution. 

Let us study the limit disribution $S$. As @A.S. wrote, its cumulants are $(1-2^{-k})^{-1}$. 
Further, let $0.\alpha_1\alpha_2\alpha_3\dots$ be the binary form of fractional part of $S$. Then the sequence $\{\alpha_n,n\ge 1\}$ of digits is stationary (in fact, this is the case for any random variable of the form $\sum_{k=0}^\infty \zeta_k 2^{-k}$ with iid integer-valued variables $\zeta_k$). It is possible to "identify" the marginal distribution of digits, that is, the probability $p = P(\alpha_1 = 1)$. Define for $k\ge 0$
$$
E_k = 1 - 2e^{-1}\sum_{n=0}^\infty \frac{1}{(2^k (2n+1))!}.
$$
Then 
$$
p = \frac12\left(1-\prod_{k=0}^\infty E_k\right).
$$
If $p\neq 1/2$, which seems to be the case, then the distribution of $S$ is singular. (In the unlikely case that $p=1/2$, it is still singular since the digits $\alpha_n$ are dependent.) It is also continuous, but at the moment I don't see a simple way to prove this.