# Does $S_b$ become eventually greater than $S_{b+1}$?

For a base $$b \geq 2$$, let $$S_b$$ be the sequence the $$n$$−th term of which is found by concatenating $$n$$ copies of the number $$n$$ written in base $$b$$, and then seeing the resulting string as a number in base $$b$$.

For example, given $$b=2$$, we get the sequence $$1$$, $$1010_2$$, $$111111_2$$, $$100100100100_2$$ etc, and for base ten we get $$1$$, $$22$$, $$333$$, $$4444$$, etc.

For fixed $$b$$, does the sequence $$S_b$$ become eventually greater than the sequence $$S_{b+1}$$? That is, does the $$n$$-th term of $$S_b$$ is greater than that of $$S_{b+1}$$ for all $$n$$ greater than some constant?

• As you go down the sequences $S_b$ and $S_{b+1}$, sometimes one will be larger, sometimes the other. The lead will swap infinitely many times. Commented Oct 24, 2022 at 13:14

The proposition is false. Consider $$S_2(n)$$ and $$S_3(n)$$, where $$n=3^k$$, $$k\in\mathbb{N}.$$

$$n = {1\underbrace{00\ldots 00}_{k \text{ zeros}} } (\text{ base } 3).$$ This pattern $${1\underbrace{00\ldots 00}_{k \text{ zeros}} }$$ repeated i.e. concatenated onto itself $$\ 3^k$$ times, $$\underbrace{100\ldots 00}_{1}\ \underbrace{100\ldots 00}_{2}\quad \ldots\quad \underbrace{100\ldots 00}_{3^k}\$$ in base $$3$$ is $$\ > 3^{(k+1)3^k - 1}$$.

On the other hand, $$n=2^{k\log_2(3)\ }$$ in base $$2$$ has at most $$\ \left\lceil k \log_2(3) \right\rceil\$$ digits, and so $$3^k$$ in base $$2$$, repeated $$3^k$$ times - which has at most $$\ 3^k\left\lceil k \log_2(3) \right\rceil\$$ digits - considered as a number in base 2, is: $$\ \leq 2^{\left\lceil k \log_2(3) \right\rceil\ \cdot3^k} \leq 2^{ (k \log_2(3) + 1) \cdot3^k } = 2^{ k3^k\left( 1 + \frac{1}{k\log_2(3)} \right) \log_2(3) } = 3^{ 3^k \left(k + \frac{1}{\log_2(3)}\right) },\$$ *which is $$\leq 3^{3^k (k+1) - 1}\$$ for all $$k\geq 1.\quad$$

$$*\quad$$ The fact that $$\ 3^k\left(k + \frac{1}{\log_2(3)}\right) \leq 3^k (k+1) - 1\ \forall\ k\geq 1\$$ is elementary: start by dividing both sides by $$3^k$$ and use the fact that $$\ \frac{1}{\log_2(3)} \approx 0.63$$.

• The sequence is not defined in base 10. It is defined in base $b$. You are concatenating numbers in base 10, while you should do it in base $b$. Commented Oct 24, 2022 at 22:07
• No, I'm concatenating them in base $b$. Any number can be represented in any base: the fact that I'm choosing to sometimes represent it in base $10$ because that is what is familiar doesn't detract from my argument. Unless I'm wrong, of course. Commented Oct 24, 2022 at 22:11
• So what does it mean the first inequality $$3^k > 3^{(k+1)(3^k)-1}$$ which is clearly false Commented Oct 24, 2022 at 22:12
• I meant to say, "concatenated onto itself $3^k$ times". So for example, with $k=4,$ we have: $81$ (base $10) =3^4$ (base $10) = 10000 (\text{ base } 3).$ This pattern $10000$ repeated i.e. concatenated onto itself $\ 81$ times is $10000100001000010000...10000 (\text{ base } 3)$$\ > 3^{(4+1)3^4 - 1}$ (base $10).$ Commented Oct 24, 2022 at 22:19
• You were right that I was unclear about this, and I've edited my answer accordingly. Is it clearer now? Commented Oct 24, 2022 at 22:29

There is an explicit formula for the sequence $$S_b$$: $$S_b(n)= \sum_{k=0}^n b^{(\lfloor \log_bn \rfloor +1)k} \cdot n \approx b^{n^2} \cdot n$$ From this formula it is clear that for all $$n$$ $$S_b (n) < S_{b+1}(n)$$

• $S_4(4)=17476>624=S_{5}(4)$ Commented Oct 24, 2022 at 23:47
• I think it's pretty clear that $S_b(b) > S_{b+1}(b)$ for all $b>1$, as the number of digits is twice as large, and the base on the right-hand side being 1 larger can't hope to make up for that. Commented Oct 25, 2022 at 7:26