Analyzing powers of $2$ after finding this interesting question, I found some patterns that keep me intrigued, and that can be visualized in the following table:
First Pattern:
Let us denote as $S(2^k)$ the sum of digits of the decimal expansion of $2^k$. Then, as it can be seen in the table, it seems that $S(2^n)=S(2^m)$, with $m\geq n$, only if $m-n=6$. For instance, $S(2^3)=S(2^9)$, $S(2^{10})=S(2^{16})$, $\dots$
Second Pattern:
The number of powers of 2 being digits of $2^k$ seems to follow some kind of palindromic symmetries. For instance, $\{1,3,1\}$, $\{2,3,3,1,3,3,2\}$, $\dots$
Third Pattern (related to first pattern?):
It seems that $S(2^{2n})=3n+1$, and when it is not the case, the digits of the decimal expansion of $2^{2n}$ can be arranged in subsequences such that its sum is $3n+1$. For instance, $2^{2*9}=262144$, and $2+6+2+14+4=28=3*9+1$.
It seems that $S(2^{2n+1})$ has some kind of pattern too, but I have not been able to identify it completely; only that $S(2^{2n+1})=3k+2$ and $S(2^{2m+1})-S(2^{2n+1})=3x$, with $k,x\in \mathbb Z$
I am a bit lost trying to understand why this patterns arise, so any help would be welcomed. Thanks!
By arranging the decimal expansion of a number in subsequences we can only increase the digit sum. Therefore, the third pattern cannot hold for $n \in \{16, 17, 18, 19, 33, 37, 47, 75, 76, 93, ...\}$ because the digit sum of $4^n$ is bigger than $3n + 1$.
$$4^{16} = 4294967296 \quad \to 58 > 49 = 3 \cdot 16 + 1$$
However, assuming the digits of $4^n$ are uniformly distributed the expected value of the digit sum is $9n \lg2 = 2.71n$ which is always smaller than $3n + 1$ so your pattern should hold for almost all numbers.
By arranging the decimal expansion of a number in subsequences we can only increase the digit sum by multiples of nine.
$$S(25139) = 20 \qquad \begin{array}{l} 2 + 5 + \mathbf{1}3 + 9 = 29 = S(25139) + 9 \cdot \mathbf{1} \\ 2 + \mathbf{5}1 + 3 + 9 = 65 = S(25139) + 9 \cdot \mathbf{5} \end{array}$$
Conveniently, the difference between the digit sum of $4^n$ and $3n + 1$ is always divisible by nine:
$$ S(4^n) \equiv \begin{cases} 1 \\ 4 \\ 7 \end{cases} \mod{9} \iff n \equiv \begin{cases} 0 \\ 1 \\ 2 \end{cases} \mod{3}$$ $$\implies 3n + 1 - S(4^n) \equiv 0 \mod{9}$$
So if the digit sum of $4^n$ is smaller than $3n+1$ we know the subsequence needs to start at the digit $$x(n) = \frac{3n + 1 - S(4^n)}9$$
$$\begin{array}{c|c|c|c} n & 4^n & x(n) & 3n + 1 = \\ \hline 5 & 1024 & 1 & 10 + 2 + 4 \\ 29 & 288230376151711744 & 2 & 28 + 8 + \cdots \\ 35 & 1180591620717411303424 & 4 & \cdots + 3 + 42 + 4 \end{array}$$
In some cases like $$4^{21} = 4398046511104 \qquad x(21) = 2 = 1+1 \qquad 64 = \cdots + 5 + 11 + 10 + 4$$
and for $x(n) \geq 10$ we need multiple subsequences where $x(n)$ is the sum of their inital digits
$$4^{185} \approx 240490760476040522535882813 \dots$$ $$\begin{align} x(185) = 13 & = 4 + 9 & 556 & = 2 + 40 + 4 + 90 + \cdots \\ & = 3 \cdot 4 + 1 & & = 2 + 40 + 49 + 7 + 6 + 47 + \cdots + 13 + \cdots \end{align} $$
Because the number of partitions of $x(n)$ increases superpolynomially it should be possible to find an arangement for almost all numbers.