OFFSET
1,2
COMMENTS
Repeated frequency followed by digit-indication. Repeating the frequency allows 5 to appear, in addition to 1, 2 and 3 which are already contained in Conway's original look-and-say sequence. However, 4 still does not appear.
The sequence is determined by triples of digits. The first two terms of a triple are the repeated figure and the last term is the digit.
Therefore, sequences of form xy (x != y), xxyy can never appear. A fortiori, the sequence never contains series of four identical digits, but contains series of five 3, which make appear the 5's (55 and 5). However five 5's never appear. Proof: suppose it appears for the first time in a(n)-a(n+4); because of 'five five 5' in 55555, it would imply that 55555 appears form a smaller n, which is a contradiction. By the same argument, 555 also never appear.
Also 22222 or 11111 are impossible : 22222 would imply a preceding 22yy and 11111 a preceding 1x (x != 1), but both cannot exist.
All terms end with 1 (the seed) and, except the first two, begin with 2 or 3.
EXAMPLE
The term after 331 is obtained by saying (repeating) two two 3, one one 1, which gives 223111.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Jean-Christophe Hervé, May 12 2013
STATUS
approved