

A225331


A continuous "lookandrepeat" sequence (method 2).


6



1, 1, 1, 1, 3, 3, 1, 2, 2, 3, 1, 1, 1, 2, 2, 2, 1, 1, 3, 3, 3, 1, 3, 3, 2, 2, 2, 1, 3, 3, 3, 1, 1, 1, 2, 2, 3, 3, 3, 2, 1, 1, 1, 3, 3, 3, 3, 3, 1, 2, 2, 2, 3, 3, 3, 1, 1, 2, 3, 3, 1, 5, 5, 3, 1, 1, 1, 3, 3, 2, 3, 3
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OFFSET

1,5


COMMENTS

A variant of the 'lookandrepeat' sequence A225329, without run cutoff. It describes at each step the preceding digits by repeating the frequency number.
The sequence is determined by triples of digits. The first two terms of a triple are the repeated frequency and the last term is the digit.
There are different optional rules to build such a sequence. This method 2 never considers twice the already said digits.
With this rule and seed, a(n) is always equal to 1, 2, 3 or 5, and the sequence is the simple concatenation of the lookandrepeat sequence by block A225329. This is because all blocks of A225329 begin with 2 or 3 and end with 2 and therefore, there is no possible interaction between blocks after concatenation.
It never contains series of four identical digits (except the first four one's), but contains series of five identical digits. 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.


LINKS

Table of n, a(n) for n=1..72.


EXAMPLE

a(1) = 1, then a(2) = a(3) = a(4) = 1 (one one 1). Leaving out the first 1 already said, we have now three 1, then a(5) = a(6) = 3, and a(7) = 1, etc.


CROSSREFS

Cf. A225330 (a close variant with 4's), A225329 (lookandrepeat by block), A005150 (original lookandsay), A225224, A221646, A225212 (continuous lookandsay versions).
Sequence in context: A317413 A110628 A107292 * A004550 A096836 A096995
Adjacent sequences: A225328 A225329 A225330 * A225332 A225333 A225334


KEYWORD

nonn,easy


AUTHOR

JeanChristophe HervĂ©, May 12 2013


STATUS

approved



