# How to prove that all these sets equal $\mathbb N$?

Let us take two natural numbers, for example $a_1=2$ and $a_2=7$. Multiply to obtain $a_3=2 \cdot 7=14$. Obtain $a_4$ as $a_2 \cdot a_3=98$. Then $a_5$ as $a_3 \cdot a_4=14 \cdot 98=1372$. Repeat this procedure an infinite number of times by defining $a_{n+2}=a_{n+1} \cdot a_n$ for $n \in \mathbb N$.

From three multiplications in this example we obtained these numbers $2,7,14,98,1372$.

If we take all possible subsets of digits of these numbers we obtain a set $\{1,2,3,4,7,8,9,12,13,14,17,32,37,72,98.132,137,172,372,1372\}$

I would like to know:

If we choose some two naturals $a_1$ and $a_2$ such that $a_1 \cdot a_2$ does not end in zero and take all possible subsets of digits of numbers in the set $\{a_{n+2}=a_{n+1} \cdot a_n\mid n\in \mathbb N\} \cup \{a_1\} \cup \{a_2\}$ do we have that set of all possible subsets of digits of numbers in the set $\{a_{n+2}=a_{n+1} \cdot a_n\mid n\in \mathbb N\} \cup \{a_1\} \cup \{a_2\}$ equals $\mathbb N$?

This seems to be way too trivial property of these sets, that is, that they all equal $\mathbb N$, so I expect some clever argument that settles this question of mine.

This is not a result of some research of some well-known research problems and topics but just a result of some thinking about some problem from recreational mathematics.

• You'd actually need to prohibit $a_2$ from being a power of $10$, even if $a_1$ is not. And you'd need to prohibit $a_1=2^k$, $a_2=5^k$, as well as some related values like $a_1=4$, $a_2=5$. – Greg Martin Mar 30 '18 at 19:48
• @GregMartin I made an edit that covers all those cases. Thanks for that reminder. – Shalom Mar 30 '18 at 19:59

This is an extended comment, not an answer: I suspect that much more is true, there is an $a_n$ whose decimal representation begins with any given sequence of digits!
Let us reformulate the above stronger claim in simpler terms. The sequence $A_n = \log_{10} a_n$ satisfies the Fibonacci recurrence relation $A_{n+1} = A_n + A_{n-1}$, so $A_n = \alpha p^n + \beta q^n$ for some $\alpha, \beta \in \mathbb{R}$, where $p = (1 + \sqrt{5}) / 2$ and $q = (1 - \sqrt{5}) / 2$. The constants $\alpha$ and $\beta$ are easily expressed in terms of $a_1$ and $a_2$.
Fix $k > 0$. Since $p > 1$ and $|q| < 1$, for $n$ large enough the first $k$ digits of $a_n = 10^{A_n}$ and $b_n = 10^{\alpha p^n}$ coincide. Thus, it is enough to study the latter sequence. Or, even better, the sequence $B_n = \alpha p^n \bmod 1$: the numbers $b_n$ and $10^{B_n}$ have the same decimal expansion, except that they have the decimal point at a different position.
Our claim is, therefore, equivalent to the following statement: the sequence $(B_n)$ is dense in $[0, 1)$.
The sequence $(B_n)$ is not dense if $\alpha = c + d \sqrt{5}$ for some integers $c, d$: the fractional parts of $p^n$ and $\sqrt{5} p^n$ accumulate near $0$ or $1$. I suspect that for other values of $\alpha$ the sequence $(c_n)$ is indeed dense in $[0, 1)$. That said, I have no clue how to prove this: someone more experienced in number theory and dynamical systems might be able to tell more.