Looking at number-lengths to iterate

We start S with a(1) = 1 and a(2) = 2.

S is always extended with the integer k by obeying these two simple rules:

1/ if the last two terms of S have the same length, they are concatenated to form k;

2/ else the longest term is divided by the shortest one and k is the integer part.

Example

As the last two terms (1 and 2) of S have the same length, they are concatenated to form a(3):

S = 1, 2, 12, …

As the last two terms of S have not the same length, we divide the longest one by the shortest one (this is 12/2 = 6) and extend S with 6:

S = 1, 2, 12, 6, …

As the last two terms of S have not the same length, we make 12/6 = 2 and extend S accordingly:

S = 1, 2, 12, 6, 2, …

As the last two terms (6 and 2) of S have the same length, we extend S with the concatenation:

S = 1, 2, 12, 6, 2, 62, …

As the last two terms of S have not the same length, we make 62/2 = 31 and extend S:

S = 1, 2, 12, 6, 2, 62, 31, …

As the last two terms (62 and 31) of S have the same length, we extend S with the concatenation:

S = 1, 2, 12, 6, 2, 62, 31, 6231, …

As the last two terms of S have not the same length, we make 6231/31 = 201:

S = 1, 2, 12, 6, 2, 62, 31, 6231, 201, …

As the last two terms of S have not the same length, we make 6231/201 = 31:

S = 1, 2, 12, 6, 2, 62, 31, 6231, 201, 31, …

As the last two terms of S have not the same length, we make 201/31 = 6.4838709… and keep only the integer part 6 to extend S:

S = 1, 2, 12, 6, 2, 62, 31, 6231, 201, 31, 6, …

As the last two terms of S have not the same length, we make 31/6 = 5.1666666… and keep 5:

S = 1, 2, 12, 6, 2, 62, 31, 6231, 201, 31, 6, 5, …

As the last two terms (6 and 5) have the same length, we concatenate:

S = 1, 2, 12, 6, 2, 62, 31, 6231, 201, 31, 6, 5, 65, …

Etc.

I get (with the usual caveats due to my hand calculations):

S = 1, 2, 12, 6, 2, 62, 31, 6231, 201, 31, 6, 5, 65, 13, 6513, 501, 13, 38, 1338, 35, 38, 3538, 93, 38, 9338, 245, 38, 6, 6, 66, 11, 6611, 601, 11, 54, 1154, 21, 54, 2154, 39, 55, 3955, 71, 55, 7155, 130, 55, 2, 27, 13, 2713, 208, 13, 16, 1316, 82, 16, 8216, 513, 16, 32, 1632, 51, 32, 5132, 160, 32, 5, 6, 56, 9, 6, 96, 16, 9616, 601, 16, 37, 1637, …

Question

Will S enter into a loop? Probably yes – but when?

And if S does, what about the seq T starting with a(1) = 1 and a(2) = 3?

Note that the May, 2024 seq [starting with a(1) = 5 and a(2) = 2024] converges quickly with the above S:

M^ay₂₀₂₄ = 5, 2024, 404, 5, 80, 16, 8016, 501, 16, 31, 1631, 52, 31, 5231, 168, 31, 5, 6, 56, ...

____________________

Edwin Clark was quick to spot the loop in S (this was confirmed by Giorgos Kalogeropoulos – who found the same loop with another start for S – and much more, see below):

>Here you can see when it starts looping:

> S = 1, 2, 12, 6, 2, 62, 31, 6231, 201, 31, 6, 5, 65, 13, 6513, 501, 13, 38, 1338, 35, 38, 3538, 93, 38, 9338, 245, 38, 6, 6, 66, 11, 6611, 601, 11, 54, 1154, 21, 54, 2154, 39, 55, 3955, 71, 55, 7155, 130, 55, 2, 27, 13, 2713, 208, 13, 16, 1316, 82, 16, 8216, 513, 16, 32, 1632, 51, 32, 5132, 160, 32, 5, 6, 56, 9, 6, 96, 16, 9616, 601, 16, 37, 1637, 44, 37, 4437, 119, 37, 3, 12, 4, 3, 43, 14, 4314, 308, 14, 22, 1422, 64, 22, 6422, 291, 22, 13, 2213, 170, 13, 13, 1313, 101, 13, 7, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, ...

> For a(1)=1 and a(2)=3 we have

1,3,13,4,3,43,14,4314,308,14,22,1422,64,22,6422,291,22,13,2213,170,13,13,1313,101,13,7,1,71,71,7171,101,71,1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, 1, 71, 71, 7171, 101, 71, ...

> same loop!

> For a(1)=2 and a(2)=3 we have the loop:

2,3,23,7,3,73,24,7324,305,24,12,2412,201,12,16,1216,76,16,7616,476,16,29,1629,56,29,5629,194,29,6,4,64,16,6416,401,16,25,1625,65,25,6525,261,25,10,2510,251,10,25,1025,41,25,4125,165,25,6,4,64,16,6416,401,16,25,1625,65,25,6525,261,25,10,2510,251,10,25,1025,41,25,4125,165,25,6,4,64,16,6416,401,16,25,1625,65,25,6525,261,25,10,2510,251,10,25,1025,41,25,4125,165,25,6,4,64,16,6416,401,16,25,1625,65,25,6525,261,25,10,2510,251,10,25,1025,41,25,4125,165,25,...

> or a(1)=1 and a(2)=4 we have the loop:

1,4,14,3,4,34,8,4,84,21,8421,401,21,19,2119,111,19,5,3,53,17,5317,312,17,18,1718,95,18,9518,528,18,29,1829,63,29,6329,218,29,7,4,74,18,7418,412,18,22,1822,82,22,8222,373,22,16,2216,138,16,8,2,82,41,8241,201,41,4,10,2,5,25,5,5,55,11,5511,501,11,45,1145,25,45,2545,56,45,5645,125,45,2,22,11,2211,201,11,18,1118,62,18,6218,345,18,19,1819,95,19,9519,501,19,26,1926,74,26,7426,285,26,10,2610,261,10,26,1026,39,26,3926,151,26,5,5,55,11,5511,501,11,45,1145,25,45,2545,56,45,5645,125,45,2,22,11,2211,201,11,18,1118,62,18,6218,345,18,19,1819,95,19,9519,501,19,26,1926,74,26,7426,285,26,10,2610,261,10,26,1026,39,26,3926,151,26,5,5,55,11,5511,501,11,45,1145,25,45,2545,56,45,5645,125,45,2,22,11,2211,201,11,18,1118,62,18,6218,345,18,19,1819,95,19,9519,501,19,26,1926,74,26,7426,285,26,10,2610,261,10,26,1026,39,26,3926,151,26,5,5,55,11,5511,501,11,45,1145,25,45,2545,56,45,5645,125,45,2,22,11,2211,201,11,18,1118,62,18,6218,345,18,19,1819,95,19,9519,501,19,26,1926,74,26,7426,285,26,10,2610,261,10,26,1026,39,26,3926,151,26,...

> I hope I’ll find some time later to investigate further your questions, many things can happen with this seq!

ÉA

> Indeed – many thanks Giorgos (and Edwin)!

> And what about this old trick that guarantees an infinite such seq:

« When a new term is the start of a loop, don’t use it but extend instead the seq with the smallest term not yet present in the seq »!

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Next day (May 30th) update

Hans Havermann has posted a very interesting remark about those loops:

> The (1, 71, 71, 7171, 101, 71) loop is an example of

(1, x, x, x<>x, 101, x), 51 <= x <= 99.

It is also illustrative of the following infinite extensions:

(1, x, x, x<>x, 1001, x), 501 <= x <= 999

(1, x, x, x<>x, 10001, x), 5001 <= x <= 9999

...

Giorgos found examples for these two:

(4, 64, 16, 6416, 401, 16, 25, 1625, 65, 25, 6525, 261, 25, 10, 2510, 251, 10, 25, 1025, 41, 25, 4125, 165, 25, 6)

and

(2, 22, 11, 2211, 201, 11, 18, 1118, 62, 18, 6218, 345, 18, 19, 1819, 95, 19, 9519, 501, 19, 26, 1926, 74, 26, 7426, 285, 26, 10, 2610, 261, 10, 26, 1026, 39, 26, 3926, 151, 26, 5, 5, 55, 11, 5511, 501, 11, 45, 1145, 25, 45, 2545, 56, 45, 5645, 125, 45).

In a quick search (not meant to be exhaustive), that leaves uncharted (thus far) these three:

(10, 10, 1010, 101)

(3, 93, 31, 9331, 301, 31, 9)

(10, 21, 1021, 48, 21, 4821, 229, 21, 10, 2110, 211)

ÉA

> Many thanks, Hans!

____________________

Next update, same day – Giorgos:

> I was trying to make a fractal out of this seq for you and I think I got it. Here is how it works:

We always take a(1)=1 and we use different values n for a(2).

Then, for each n we mark the position of the first occurrence of a pair of duplicate terms

in the example, for n=1, we get the seq a(1)=1 and a(2)=1.

n=1:

1,1,11,11,1111,101,11,9,1,91,91,9191,101,91,1,91,91,9191,101,91,1,91,91,9191,101,91,1,91,91,9191,101,91,1,91,91,9191,101,91,1,91,91,9191,101,91,1,91,91,9191,101,91...

We see that the first duplicate happens at positions 15 and 16 and we will call it just 16.

So, for different n we get the following fractal F seq:

F = 16,117,34,125,65,129,129,129,129,7,15,15,15,43,120,120,114,114,114,65,68,68,68,139,139,129,129,129,129,79,79,79,79,77,77,77,77,77,77,90,90,20,20,63,62,60,60,53,53,71,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,12,23,23,23,23,23,23,23,23,23,23,23,51,51,51,51,51,51,51,51,128,128,128,128,128,128,134,134,134,134,134,134,134,134,134,134,134,134,134,134,69,69,69,69,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,141,115,115,115,115,115,115,115,115,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,67,109,...

Here are the first 1000 terms