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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:

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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!

Juan Moreno
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    As a cheap remark: Note that $S(m)\equiv m\pmod 9$ and that the order of $2\pmod 9$ is $6$. Thus $S(2^{m+6})\equiv S(2^m)\pmod 9$ at least. If $S(2^{m+k})=S(2^m)$ then $k$ is a multiple of $6$. – lulu Mar 16 '22 at 10:20
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    The pair $[128175, 129675]$ exhibits $m-n=1500$ and the digitsum is $174\ 518$ in both cases. – Peter Mar 16 '22 at 16:45
  • @lulu thanks! Really insightful comment! – Juan Moreno Mar 16 '22 at 18:10
  • @Peter great counterexample to the first pattern. Is it the first one that you have found? A bit hard to calculate... – Juan Moreno Mar 16 '22 at 18:11
  • It is the smallest with EXACT difference $1500$ , but need not be the smallest with difference larger than $1500$ – Peter Mar 16 '22 at 18:12
  • In fact, this beats my counterexample : $[65215, 66859, 1644]$ , digitsum in both case is $89\ 507$. In fact significantly more extreme. I guess that the palindrome pattern also eventually breaks away. – Peter Mar 16 '22 at 18:18
  • My current record for the difference occurs here : $[233725, 237637, 3912]$ – Peter Mar 16 '22 at 18:25
  • PARI/GP can easily calculate the digitsums. The last pair has common sum $318\ 593$ – Peter Mar 16 '22 at 18:29
  • @Peter I also guess that the palindrome pattern eventually breaks away; too amazing to be true! But I still wonder what does generate it – Juan Moreno Mar 16 '22 at 18:49

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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.

Christian
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