Call a number "holy" if it contains no $666$ in its decimal expansion. Are there infinitely many holy powers of $2$?

We call a number "holy" if it contains no $666$ in its decimal expansion, and "unholy" otherwise. For instance, $12366621$ and $666609$ are unholy, while $7777$ and $66166266366$ are holy.

Question: Is the set $$\{2^n \ | \ n \in \mathbb N, 2^n \text{ is holy}\}$$ infinite?

Of course, tons of similar questions can be asked by changing the number $666$, the base $2$, and the base for extension (we asked for decimal, so the default was $10$). I do not feel that I am the first one asking this, and I appreciate it if someone gives me references if applicable.

But my thought is the following:

Conjecture: No.

I will share my reasoning at the end of the post, but let us first see some facts:

Smallest unholy instances: $$ \begin{aligned} 2^{157} &= 182687704\color{magenta}{666}362864775460604089535377456991567872\\ 2^{192} &= 6277101735386680763835789423207\color{magenta}{666}416102355444464034512896 \end{aligned} $$

Then, we witnessed a cluster of unholy powers: $2^{218}, 2^{220}, 2^{222}, 2^{224}, 2^{226}$, and then kept holy for a while, until we hit the unholy $2^{243}$.

Largest holy instances: I did not throw in a lot of CPU time to pursue holy numbers, nor did I try hard enough to optimize my programs, but among the $3715$ holy powers of $2$, the largest of them are $$2^{25357}, 2^{25896}, 2^{26051}, 2^{26667}, 2^{29784}.$$

I tested up to around $2^{110000}$, but that is all I got. It probably will be reasonable for an average computer to test up to say $2^{10^6}$ or $2^{10^7}$, but I will be surprised to see a new holy number.

Statistics: For an integer $n$, let $H(n)$ be the number of holy powers of $2$ up to $2^n$.

n      | H(n)     || n      | H(n)     || n      | H(n)
 1000  |  875     || 11000  | 3567     || 21000  | 3700
 2000  | 1560     || 12000  | 3602     || 22000  | 3703
 3000  | 2059     || 13000  | 3621     || 23000  | 3705
 4000  | 2442     || 14000  | 3645     || 24000  | 3707
 5000  | 2747     || 15000  | 3655     || 25000  | 3709
 6000  | 2984     || 16000  | 3670     || 26000  | 3712
 7000  | 3171     || 17000  | 3682     || 27000  | 3714
 8000  | 3332     || 18000  | 3689     || 28000  | 3714
 9000  | 3440     || 19000  | 3693     || 29000  | 3714
10000  | 3514     || 20000  | 3695     || 30000  | 3715

A plot of this:

The heuristics of the conjecture:

This is definitely not close to a proof at all, and I still hope if rigorous arguments exists:

The idea is that we want to estimate, for an integer $n$, the probability $P(n)$ that $2^n$ is holy, and then compute $\sum_{n=1}^\infty P(n)$.

We know that $2^n$ has $O(n\ln 2)$ decimal digits, so there are $O(n\ln 2)$ groups of three. For each group there is a $1-10^{-3}$ chance to be not $666$, so very roughly $$ P(n) = (1-10^{-3})^{n\ln 2} \approx e^{-10^{-3}\ n\ln 2}. $$

And note that $$ \sum_{n=1}^\infty P(n) \approx \int_{n=0}^\infty e^{-10^{-3}\ x\ln 2} dx < \infty. $$

The red "estimation line" in the figure above follows from this integral.

Of course, one can easily argue the properness of the heuristic above:

• The distribution of the digits close to the left are not uniform; they are affected by the growth of logarithmic functions.
• The distribution of the digits close to the right are not uniform; they are affected by the pattern of $2^n \pmod{10^k}$.
• $P(n)$ and $P(n+i)$ are not independent, partially because of the awful choice of the number $666$: $6\cdots 6 \times 2^2 = 26\cdots 64$.

Any thoughts are appreciated.

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