Finitude of an alternative of the Look and Say serie?

Question

Consider the following alternative of the Look and Say sequence (OEIS A005150):

$u_0=N>0$

$u_{n+1}$ is the number of 1s then the number of 2s then ... then the number of 9s in $u_n$.

Example : $u_0=5,u_1=15,u_2=1115,u_3=3115,u_4=211315,u_5=31121315,u_6=41122315,u_7=3122131415,u_8=41122315,u_9=3132132415,u_{10}=3122331415,u_{11}=3122331415=u_{10},\dots$.

Notice how the sequence starts repeating after $u_{10}$.

Using a little python program, I found the following (see the data below for extra information):

• for $1\le u_0\le 45$, the sequence repeats itself with a step of one (repetition of one term)

• for $46\le u_0\le 56$, the sequence repeats itself with a step of two or less (repetition of two terms alternatively maximum)

• for $57\le u_0\le 5988$, the sequence repeats itself with a step of three or less (repetition of three terms alternatively maximum)

• I didn't check for higher numbers.

For any $u_0\ge1$, does the sequence have a finite number of images ?

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Data for $1\le u_0\le 45$ (format : $u_0$,final number, first step where it repeats):

(1, '21322314', 13)

(2, '21322314', 12)

(3, '21322314', 13)

(4, '21322314', 9)

(5, '3122331415', 11)

(6, '3122331416', 11)

(7, '3122331417', 11)

(8, '3122331418', 11)

(9, '3122331419', 11)

(10, '21322314', 13)

(11, '21322314', 12)

(12, '21322314', 11)

(13, '21322314', 12)

(14, '21322314', 8)

(15, '3122331415', 10

(16, '3122331416', 10

(17, '3122331417', 10

(18, '3122331418', 10

(19, '3122331419', 10

(20, '21322314', 12)

(21, '21322314', 11)

(22, '22', 1)

(23, '21322314', 10)

(24, '31123314', 7)

(25, '31123315', 7)

(26, '31123316', 7)

(27, '31123317', 7)

(28, '31123318', 7)

(29, '31123319', 7)

(30, '21322314', 13)

(31, '21322314', 12)

(32, '21322314', 10)

(33, '21322314', 11)

(34, '21322314', 7)

(35, '3122331415', 9)

(36, '3122331416', 9)

(37, '3122331417', 9)

(38, '3122331418', 9)

(39, '3122331419', 9)

(40, '21322314', 9)

(41, '21322314', 8)

(42, '31123314', 7)

(43, '21322314', 7)

(44, '31123314', 8)

(45, '3122331415', 9)

Answer 1

Yes, the sequence always ends up at a cycle of finite length.

It is fairly easy to see that every number with 22 or more digits becomes strictly shorter by your modified look-and-say rule, so no matter which you start at, sooner or later you will end up below $10^{22}$, and then you will reach a number you have been at before after at most $10^{22}$ more steps.

Answer 2

If multi-digit numbers are allowed, the game still ends in a cycle every time. This was proven in 1994 apparently: https://www.jstor.org/stable/2975323?seq=1#metadata_info_tab_contents

Interestingly, all the cycles I found by computer search fall into 1 of 7 families: $$1)\quad22$$ $$2)\quad2132231a\quad\text{with }a\ge4$$ $$3)\quad 3112331a\quad\text{with }a\ge4$$ $$4)\quad 3122331a1b\quad\text{with }a,b\ge4\text{ and }a\not=b$$ $$5)\quad 314213241a1b\quad\rightarrow\quad412223241a1b\quad\rightarrow\quad...\quad\text{with }a,b\ge5\text{ and }a\not=b$$ $$6)\quad 51222314251a1b\quad\rightarrow\quad41421314251a1b\quad\rightarrow\quad51221334151a1b\quad\rightarrow\quad...$$ $$\text{with }a,b\ge6\text{ and }a\not=b$$ $$7)\quad (n-1)14213141(n-1)2n1a_1...1a_{n-3}\quad\rightarrow\quad n12213242(n-1)1n1a_1...1a_{n-3}\quad\rightarrow\quad...\quad\text{with distinct }a_i\not\in\{1,2,4,n-1,n\}\text{ and }n\ge 6$$ That last one is pretty ugly so lemme give an example. Choosing $n=7$ and $a_1,a_2,a_3,a_4=3,8,10,384$ for examples gives us a two-cycle: $$61423141627181(10)1(384)$$ $$\downarrow$$ $$712213242617181(10)1(384)$$ $$\downarrow$$ $$...$$