Inspired by Does this sequence reach infinity?, I propose the following generalization:
$F_b(n)$ replaces consecutive repeating digits of $n$ in base $b$, with just one digit.
Find smallest $m\in\mathbb N$ such that sequence $a(0)=1, a(k)=F_b(m\cdot a(k-1))\to\infty$.
For example, $F_2(17)=F_2(2^4+1)=F_2(10001_2)=101_2=2^2+1=5$.
E.g. $a(k)$ for $b=10,m=2$ does not diverge to $\infty$, because it loops $a(68)=a(4)=16$.
(b=10, m=2) => a(k) = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 6536, 13072, 2614, 528, 1056, 212, 424, 848, 1696, 392, 784, 1568, 3136, 6272, 1254, 2508, 5016, 1032, 2064, 4128, 8256, 16512, 3024, 6048, 12096, 24192, 48384, 96768, 193536, 387072, 7414, 14828, 29656, 59312, 18624, 37248, 7496, 1492, 2984, 5968, 1936, 3872, 74, 148, 296, 592, 184, 368, 736, 1472, 294, 58, 16, ...
I couldn't find such $m$ for base $b=10$, so I've tried looking at smaller number bases $b$.
Base $2$
The $m=17=10001_2$ diverges in base $2$, because it follows the following increasing pattern:
$$ a(k)=F_2(10\color{red}{11}\color{blue}{\overline{01}}\dots\color{blue}{\overline{01}}\color{red}{00}101_2) $$
after $k\ge 2$. By applying $F_2$ and evaluating $a(k)$, we see $a(k)=4 a(k-1) + 1$ diverges to $\infty$.
That is, in this case, the "overlined blue digit substrings $\color{blue}{\overline{01}}$" repeats $k-2$ times.
The proof is inductive. Notice $m=17=10001_2$ and assume $a(k)=1010(10)_{k-2}101_2$. Then
$$\begin{align} a(k+1) &= F_2(m\cdot a(k)) \\ &= F_2(10001_2\cdot 1010(10)_{k-2}101_2) \\ &= F_2(1010(10)_{k-2}1010000_2 + 1010(10)_{k-4}1010101_2) \\ &= F_2(1011(01)_{k-2}0100101_2) \\ &= 101(01)_{k-2}010101_2 \\ &= 1010(10)_{k-2}10101_2 \\ &= 1010(10)_{k-1}101_2 \end{align}$$
as expected, and the base case is $$a(0)=1, a(1)=5\implies a(2)=85=1010101_2=1010(10)_{0}101_2.$$
Base $3$
The $m=14=112_3$ diverges in base $3$, because it cycles over $12$ increasing patterns:
$$\begin{align} a(12k+0) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}102\color{red}{00}20101_3)\\ a(12k+1) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}\color{red}{22}0\color{red}{22}012_3) \\ a(12k+2) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}10121_3)\\ a(12k+3) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}1020120120\color{red}{22}_3)\\ a(12k+4) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}102\color{red}{00}20101_3)\\ a(12k+5) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}\color{red}{11}0\color{red}{22}012_3)\\ a(12k+6) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}10121_3)\\ a(12k+7) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}1020120120\color{red}{22}_3)\\ a(12k+8) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}102\color{red}{00}20101_3)\\ a(12k+9) &= F_3(10\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}\color{red}{11}0\color{red}{22}012_3)\\ a(12k+10) &= F_3(120\color{red}{22}\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}10121_3)\\ a(12k+11) &= F_3(\color{red}{22}01202\color{blue}{\overline{10201202}}\dots\color{blue}{\overline{10201202}}1020120120\color{red}{22}_3)\\ \end{align}$$
after $(12k+0)\ge 24$, where all patterns diverge to $\infty$. This can be proven by induction.
Base $4$
The $m=2046=133332_4$ diverges in base $4$, because it cycles over $2$ increasing patterns:
$$\begin{align} a(2k+0)&=F_3(323\color{red}{00}\color{blue}{\overline{103230}}\dots\color{blue}{\overline{103230}}10\color{red}{33}2301\color{red}{00}_4)\\ a(2k+1)&=F_3(1312\color{red}{00}\color{blue}{\overline{213120}}\dots\color{blue}{\overline{213120}}21312\color{red}{00}21320_4) \end{align}$$
after $(2k+0)\ge 4$, where both patterns diverge to $\infty$. This can be proven by induction.
I haven't found such $m$'s for $b\ge 5$, yet.
Since finding the smallest $m$ for a given base $b$ looks hard, I am asking just about the existence:
Does for every number base $b$, there exists at least one $m$ such that $a(k)\to\infty$?
That also might be a hard question. Can we at least prove or disprove this for small bases $b\le 10$ ?
E.g. for $b=5$, all $m\le 4000$ get stuck in a loop (do not diverge to $\infty$), so far. Does such $m$ exist?
from Hot Weekly Questions - Mathematics Stack Exchange
Vepir
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