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Sequence: ABC; 1A1B1C; 111A111B111C; 311A311B311C; ...

What is the 6th term of the sequence?

a) 113121212A333311112B123112212C
b) 221133112A123212133B332211332C
c) 321321321A321321321B321321321C
d) 123123122A131221132B123123123C
e) 111312211A111312211B111312211C

I have doubt about this sequence... I didn't understand the sequence logic. any hint?

How to solve this problem?

Lord_Farin
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Victor Lellis
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    It seems to be a variant of the Look-and-say sequence. – Andrés E. Caicedo Oct 02 '13 at 22:51
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    (You begin with one A, one B, one C. The second term in the sequence tells you that this is what you had. This second term consists of one 1, one A, one 1, one B, one 1, and one C. Etc.) – Andrés E. Caicedo Oct 02 '13 at 22:52
  • The correct answer is c... ABC; 1A1B1C; 111A111B111C; 311A311B311C; 321A321B321C; 321321321A321321321B321321321C. – Victor Lellis Oct 02 '13 at 23:02
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    Victor, if you agree with the suggestion I'm giving of how the sequence is formed, your fifth and sixth terms are incorrect, the fourth terms begins with one 3, so the fifth term should begin 13... – Andrés E. Caicedo Oct 02 '13 at 23:09
  • Andres has given the useful answer, so I will give some useless (but also correct) answers: (1) nothing; the sequence has no 6th term. (2) whatever you want it to be. – Trevor Wilson Oct 02 '13 at 23:23
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    Considering the variant of the Look-and-say sequence, the answer is e... but without this consideration and using another algorithm can be whatever answer. – Victor Lellis Oct 02 '13 at 23:30
  • @TrevorWilson :-) But I addressed this as well! (Well, not the possibility that there is no sixth term, because "sequence" for me means with domain $\mathbb N$ -- unless otherwise specified.) – Andrés E. Caicedo Oct 03 '13 at 00:27
  • @Andres Ah, you are right. I guess I wanted to say it more directly (or obnoxiously.) – Trevor Wilson Oct 03 '13 at 00:33
  • @VictorLellis This question is not about logic in the mathematical sense. If anything, it is about pattern recognition. Please do not revert it again. – Lord_Farin Oct 03 '13 at 09:31
  • For those who would say that it is off-topic, please look at this page: http://en.wikipedia.org/wiki/Look-and-say_sequence Such sequences are studied as part of mathematics and so I think that it is reasonable to view this question as on topic. – Tom Oldfield Oct 03 '13 at 09:58

1 Answers1

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This seems to be an example of the Look-and-say sequence, which is in itself interesting; its variant, the Kolkoski sequence, leads to several difficult problems.

Specifically, each term of the sequence after the initial one describes the previous term of the sequence by listing the number of times each symbol appears in the term. Thus, since the first term in the sequence is ABC, the next term should tell us that we have one A, one B, and one C, and is therefore 1A1B1C. This term has one 1, one A, one 1, one B, one 1, and one C, so the third term of the sequence is 111A111B111C. Continuing this way, we have 311A311B311C, 13211A13211B13211C, and 111312211A111312211B111312211C, so the answer appears to be option e).

That being said, as it is always the case with these problems, there is nowhere near enough information to actually answer the question in a way that is mathematically justifiable, and any term whatsoever is the sixth term of a sequence that begins as the one you listed. This, of course, is never considered by whoever asks these questions.

Andrés E. Caicedo
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