Problems & Puzzles:
Look and say sequence
For sure you know what is the look & say sequence. See
Only three primes are known
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
Related with this last question, I got an unexpected
output: analyzing up to the
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
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
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
Also, the look and say conditions implie
that in a configuration ...aba... the digit b cannot be a
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...
The string ...313... is not possible
; ...313... should come from ...111xxx... ,
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,