Problems & Puzzles: Puzzles

Puzzle 657 Look and say sequence

For sure you know what is the look & say sequence. See these A, B & C, references.

Only three primes are known (Oct. 2012) in this sequence: 11, 312211  13112221

Q1- Find the fourth prime.

Some digital properties are already known. For example: No digits different from 1, 2 & 3 are contained in any term. This implies that there are not substrings of four or more of repeated digits ("...xxxx..." for x= 1, 2 or 3) in any term.

Therefore, I started asking myself about of substrings of three or less digits. Specifically I wonder to know this: is any combination of three or less digits found as substring in some terms?

Related with this last question, I got an unexpected output: analyzing up to the 25 terms easily available in the web (around 5000 total digits in the 25 terms) the following three digits combinations were absent: "333", "323", "313" & "233".

I must say that I was unable to find references to these four absent-combinations, so perhaps going further my four claims above are wrong.

Q2. Can you check this absent-combinations in larger terms of the L&S sequence, or exhibit a reason for the absences?

Contributions came from Jahangeer & Farideh Firoozbakht and Emmanuel Vantieghem-


 Jahangeer & Farideh Firoozbakht wrote:

Between the first 46 terms of the sequence A005150 (look & say sequence)there exist only three prime terms. A005150(47) has 403966 digits and we don't know this term is prime or composite. So next prime term of the sequence has more than 403965 digits...
A005150(47) has 403966 digits and for Mathemtica it is too large to determine whether it is composite or probable-prime. This test was done about few hours.


Emmanuel wrote:

Q1 : I could not find another prime among the first 46 numbers of the sequence.

Q2.  The digit combinations 333, 323, 313, 233 never occur !

Any number  m  > 1  in the sequence is composed of  two kinds of digits : counting  digits and counted digits.
For instance, in the 9th number m =  31131211131221, the counting digits are underlined. The first digit (3) is a counting digit ; it counts the number 1 (= the counted digit) of the 8th number n = 1113213211. The next digit of m is a counted digit, the next one is a counting digit, and so on.
Thus, the counting digits occur alternatively.
Also, the look and say conditions implie that in a configuration ...aba... the digit b cannot be a counting digit.

Thus, ...333... must be ...333... and should come from ...333xxx... But then either 3 or x is a counting digit which is impossible.

The string ...323... is not possible ; ...323... should come from ...222xxx... , impossible.
The string ...313... is not possible ; ...313... should come from ...111xxx... , impossible.
The string ...233... is not possible ; ...233... should come from ...x33yyy... which is only possible as ...x33yyy... , which should come from ...333... , impossible.
With the same argumentation we can prove that the following digit combinations are impossible :
1111, 1212, 1233, 1313, 1323, 1333, 2121, 2222, 2223, 2232, 2233, 2313, 2323, 2331, 2332, 2333, 3131, 3132, 3133, 3231, 3232, 3233, 3313, 3323, 3331, 3332, 3333




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