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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?

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closed as off-topic by mrf, Trevor Wilson, dfeuer, Rick Decker, Vedran Šego Oct 3 '13 at 0:42

This question appears to be off-topic. The users who voted to close gave this specific reason:

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3  
It seems to be a variant of the Look-and-say sequence. –  Andres Caicedo Oct 2 '13 at 22:51
1  
(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.) –  Andres Caicedo Oct 2 '13 at 22:52
    
The correct answer is c... ABC; 1A1B1C; 111A111B111C; 311A311B311C; 321A321B321C; 321321321A321321321B321321321C. –  Victor Lellis Oct 2 '13 at 23:02
1  
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... –  Andres Caicedo Oct 2 '13 at 23:09
1  
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 2 '13 at 23:30

1 Answer 1

up vote 5 down vote accepted

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.

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