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Here's a part of the sequence:


The letters are the missing members (there are 5 missing members) and A is the first member of the sequence. The last member presented is not the end of the sequence, it can go on forever.

The sequence begins with a single integer. Can you tell which one? Simplest solution takes the cake.


marked as duplicate by Rand al'Thor, xnor, A E, Tryth, Aza Dec 29 '14 at 2:43

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Bonus points:

The next in the series:


Unless I made a typo, I'm fairly certain this is all correct. It's the look and say sequence, where the next number describes the last;


11 (there's 1 1)

21 (there's 2 1s)

1211 (there's 1 2 and 1 1s)

So in other words, 3 is the first integer.

  • $\begingroup$ Just because you used bonus points! (both tell me you answered 6 hours ago) XD $\endgroup$ – Francisco Grosso Oct 4 '14 at 17:26
  • $\begingroup$ Follow up question: Does this never have a digit >3? Why? $\endgroup$ – ThePopMachine Oct 6 '14 at 17:43
  • $\begingroup$ @Thepop No. I forget why, but you cannot reach 4, I've brute forced the first 50 numbers of the sequence so... $\endgroup$ – warspyking Oct 6 '14 at 18:05
  • 1
    $\begingroup$ Oh, I see why. Each sequence alternates "quantity" digits and "literal" digits. You can't get two consecutive "literal" digits that are the same because then it would have been rendered as a larger quantity value. E.g. you would never get the sequence "two 2's, two 2's" -> 2222. It would have been 42. So this can never happen $\endgroup$ – ThePopMachine Oct 6 '14 at 18:10
  • 3
  • 13
  • 1113
  • 3113
  • 132113
  • 1113122113
  • 311311222113

Each value is the encoding of the previous, so starting with 3, the next will be 13, or 1 instance of the number 3. The next will be 1113, one instance of 1, one instance of 3. From here you have 3113, three 1s, one 3, etc.

EDIT: D'oh, too slow - too much messing with registration and impossible captchas :)


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