3
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I follow a pattern.

What am I?

It appears to me that just the above two sentences are not enough. I should add more about what I've already tried to solve it. But I already solved it, and it wouldn't be fun if I told more. [this is a rant, not a clue]

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    $\begingroup$ I feel like this is the epitome of a "too broad" candidate. I personally will never vote to close a question for being too broad, but I am positive this question will meet that fate unless you add anything to make it unique. $\endgroup$ Commented Jan 7, 2016 at 20:44
  • $\begingroup$ @question_asker It may, but if someone discoveres the answer before closing, then people will appreciate the beauty of the puzzle. There is also a hint in the tags. $\endgroup$
    – wythagoras
    Commented Jan 7, 2016 at 20:46
  • $\begingroup$ And the title is a big, big hint of course. $\endgroup$
    – wythagoras
    Commented Jan 7, 2016 at 20:52
  • $\begingroup$ This is (for want of a better term) a meta-puzzle since it is a puzzle 'about' a series of puzzles by the OP. I'm not sure whether this type of puzzle is well-accepted on this site, but I suppose it does form a novel category. I'm willing to vote to re-open if you add a 'meta-puzzle' tag to the puzzle. (The description for the meta-puzzle tag should be along the lines of "puzzle about puzzles".) $\endgroup$
    – Lawrence
    Commented Jan 15, 2016 at 3:29

4 Answers 4

5
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I think the answer is

51

because

The last two riddles of OP were 'the circle of three' and 'the circle of two' having answers 'two' and 'Mega'. So according to the pattern the next had to be 'the circle of Mega' whose answer should be the name of the next puzzle by OP which is 'Circle of 51'.

Great Series.

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  • $\begingroup$ So close. But there has to be a reason that there is a fourth puzzle. $\endgroup$
    – wythagoras
    Commented Jan 7, 2016 at 20:54
  • $\begingroup$ Also, without your first title I never would have thought of the other puzzles, so I'll have to thank you for that. $\endgroup$
    – wythagoras
    Commented Jan 7, 2016 at 20:57
  • $\begingroup$ @wythagoras well the fourth puzzle came after this puzzle. So maybe the pattern followed till this puzzle only. And i was just going to typing to ask you to thank me.. $\endgroup$
    – manshu
    Commented Jan 7, 2016 at 20:58
  • $\begingroup$ You won't need the fifth puzzle to solve this one. $\endgroup$
    – wythagoras
    Commented Jan 7, 2016 at 21:02
  • $\begingroup$ @wythagoras edited... is it right? $\endgroup$
    – manshu
    Commented Jan 7, 2016 at 21:03
7
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Are you

a full stop (.) or "What am I?"

because you

immediately follow the words "a pattern"

I only answered this because this was tagged .

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  • $\begingroup$ No. The answer is harder. $\endgroup$
    – wythagoras
    Commented Jan 7, 2016 at 20:39
  • $\begingroup$ @wythagoras I would have thought so... $\endgroup$
    – Artyer
    Commented Jan 7, 2016 at 20:39
2
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I'll have a go at this:

Mega is, as we've discussed in previous questions, a two (2) inside of a circle.
This notation, known as Steinhaus-Moser notation, makes the circle synonymous with a pentagon

As stated in the linked article, "a number n in a triangle means nn."
'a number n in a square is equivalent to "the number n inside n triangles, which are all nested."'
'a number n in a pentagon is equivalent with "the number n inside n squares, which are all nested."'

So you could claim that, since the circle is operationally equivalent to the pentagon, the pattern is an increasing number of sides of the enclosing shape.

I don't know how to do the fancy number shit*, so if anybody wants to give me a hand there, feel free.

* but I am a font of technical terminology such as this

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1
  • $\begingroup$ Deep thought, but not the answer. $\endgroup$
    – wythagoras
    Commented Jan 7, 2016 at 21:00
2
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Are you

Moser's number

My reasoning is

your previous question asked about 2 in a circle, which is a Mega. Using the same notation, Moser's number is 2 in a polygon with Mega sides (which would be practically indistinguishable from a circle, so a circle of Mega).

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    $\begingroup$ This also explains why he said "lateral-thinking" is a hint: He is literally talking about the number of sides $\endgroup$
    – MathET
    Commented Jan 7, 2016 at 23:42

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