The New Sudoku Players' Forum

[rjamil]

Look closely at the first five rows of this growing sequence of numbers:

Row 1: 1
Row 2: 11
Row 3: 21
Row 4: 1211
Row 5: 111221

Part 1 (The Warm-up): What is the next number (Row 6) in this sequence?
Part 2 (The Hard Core Riddle): No matter how long you continue this sequence to infinity, the digits 4, 5, 6, 7, 8, and 9 will never appear. Why?

R. Jamil

[Leren]

Removed

Leren

[Leren]

I'll just give you the answer and you can judge what it's worth.

Row 5: 111221

To get Row 6 this is what you do. The first 3 digits are 1's. So what you Say is Three Ones => 31 The next two digits are both 2 so you Say Two Two's => 22. The last digit is a 1 so you Say One One => 11

Row 6: 312211

As to why digits > 3 can't appear it's because this Look and Say Sequence is based on saying two words and the digit runs gravitate toward sets of length 2. You can see that the 111 at the start of Row 5 became a shorter sequence 31.

You can read all about the Look and Say Sequence here.

Leren

[Serg]

Hi!
Interesting challenge. I tried to find "Row 6" sequence yesterday, but I could not do it. Now, when the right answer is published (I am not sure the challenge has unique solution), I'll try to prove impossibility of digits 4, ..., 9.

Let's assume the digit "4" may appear in some "row". It would imply that previous "row" should contain "NNNN" fragment. We should consider 2 cases.
Case A: the first "N" is the second term of two-digit construction "Number of digits"+"Digit". But the second and the third digits ("NN") will describe N-digit fragment of "N" digit, so we'll get 2 adjacent constructions "Number of digits"+"N" - it's prohibited.
Case B: the first "N" is the first term of two-digit construction "Number of digits"+"Digit". Then the third and the fourth digits ("NN") will describe N-digit fragment of "N" digit, so we'll get 2 adjacent constructions "Number of digits"+"N" - it's prohibited.

So, assumption "the digit "4" may appear in some "row"" is wrong. One can prove in the same manner impossibility of 5, ..., 9 digits in a "row".

Serg

[rjamil]

The Endless Morphing Sequence is simply a look-and-say sequence; a fascinating mathematical puzzle (also known as Conway's sequence) or a traditional, whole-word approach to teaching children how to read.

Part 1: Finding Row 6
This sequence does not use traditional math operations like addition or multiplication. Instead, each row literally describes the numbers in the row right above it out loud.

Row 1 is just 1.
Row 2 describes Row 1: It has "one 1", so you write 11.
Row 3 describes Row 2: It has "two 1s", so you write 21.
Row 4 describes Row 3: It has "one 2, then one 1", so you write 1211.
Row 5 describes Row 4: It has "one 1, one 2, and two 1s", so you write 111221.

Following this exact pattern, let's read Row 5 out loud: It has "three 1s, two 2s, and one 1".
Therefore, Row 6 is 312211.

Part 2: The Paradox of the Missing Digits
Why can a digit like 4 never appear, even if the sequence grows to a billion digits long?

To get a 4 in a row, the row directly above it would need to contain four identical digits in a row (for example, 1111). Let's prove why four identical digits in a row is mathematically impossible in this sequence:
Suppose a row contains 1111.
Because the sequence reads left-to-right in pairs of (quantity + digit), those four 1s would have to be read as two separate groups.The first two 1s would mean "one 1". The next two 1s would also mean "one 1".

But if the row actually had "one 1" followed immediately by "one 1", the person writing the sequence would have combined them and written "two 1s" (21) instead of 1111.

Because the sequence naturally compresses consecutive identical digits when moving from one row to the next, you can never have more than three identical digits in a row. Since you can never write 1111, 2222, or 3333, the number 4 (or anything higher) can never be born.

R. Jamil

[Serg]

Hi, Jamil!

Suppose a row contains 1111.
Because the sequence reads left-to-right in pairs of (quantity + digit), those four 1s would have to be read as two separate groups.

I think you didn't consider variant, when "four 1s would can be read as three separate groups" - the first "1" - tail of the first group, the second and third "1"s - the second group and the last "1" - beginning of the third group.

Serg

[rjamil]

I think you didn't consider variant, when "four 1s would can be read as three separate groups" - the first "1" - tail of the first group, the second and third "1"s - the second group and the last "1" - beginning of the third group.

Well, how four-1 become three separate groups? If the first one-1 would be the digit, then what shoud be its quantity (quantity + 1)? Similarly, the same applies to fourth-1, (1 + digit)?

R. Jamil

[Serg]

Hi, Jamil!

I think you didn't consider variant, when "four 1s would can be read as three separate groups" - the first "1" - tail of the first group, the second and third "1"s - the second group and the last "1" - beginning of the third group.

Well, how four-1 become three separate groups? If the first one-1 would be the digit, then what shoud be its quantity (quantity + 1)? Similarly, the same applies to fourth-1, (1 + digit)?

This is an row example, where four "1" belong to 3 separate groups: 211112 (two "1", one "1", one "2").

Serg

[rjamil]

This is an row example, where four "1" belong to 3 separate groups: 211112 (two "1", one "1", one "2").

If this is the row (definitly not first one, and should be after first five given rows) then what is/are the row immediately above this row upto fifth given row?
OR
Row number that contain this data?

This will break the necessary constraint.

R. Jamil

[Serg]

Jamil,
if you state that every "1111" sequence (if any) must participate 2 separate groups only (but not 3 groups), you must prove it.

Serg

[rjamil]

Jamil,
if you state that every "1111" sequence (if any) must participate 2 separate groups only (but not 3 groups), you must prove it.

Serg

Well, after reveiling the riddle in my above post, there is no point to divert the constraint as described in words for first five rows. Similarly, to follow the exact pattern, one must read all same consecutive digits of above row in combined (quantity + digit) format.

Let for example, as per your data, i.e., "... four "1" belong to 3 separate groups: 211112 (two "1", one "1", one "2")":

If it follows look above row and say sequence, then the above row look like:

1 + 11 + 2 = 1112

Why not it look and say; three-1s and one-2?

However, if you read same digits in breakup, then you are focusing only in second constraint, i.e., no digits between 4 to 9 used.

R. Jamil

[Leren]

You can get a 4 if Row 1 was 1111. The standard way of saying it is Four Ones => 41 in Row 2. Once 4 is there it stays there. So does the 1. In fact the last digit in Row 1 is always the last digit in every row.

That's all mentioned in the Wikipedia article in my "here" link.

I don't think it mentions what you would do if you used a slightly longer way of Saying what you see. You could say "The number of contiguous 1s I See is 4" => 14.

I've never before seen an algorithm result that depends on the precise way you Say something.

You might say the second way of saying what you See is a bit clumsy but it might be that in other languages that might be the way that a literal translation into English might turn out.

Leren

[rjamil]

You can get a 4 if Row 1 was 1111. The standard way of saying it is Four Ones => 41 in Row 2. Once 4 is there it stays there. So does the 1. In fact the last digit in Row 1 is always the last digit in every row.

This could be the another riddle if its first five rows are not same as the OP (opening post).

Similarly, the constrain that says, "Look-and-Say" must be something open statement, like for example, "look at yourself and say whatever you see in your own way."

Changing in first five given rows will change the entire riddle if it won't follow the Conway's sequence rule (quantity + digit).

However, the last one is always there, because the digit comes after the quantity.

And, I think, switching the quantity and digit won'twill effect if the rest of the constrain are same.
At a quick look, row 20 will break the rule only 1 to 3 digits are used.

R. Jamil