What's the Sequence?
30 Apr 2009 12:15 

30 Apr 2009 12:15 
© James Kanjo 2008 
Puzzle:
2, 8, 992, 1111099008, ?
Clue:
The pattern in this series is extremely difficult, so think of complex manipulation techniques to solve it.
A friend emailed me this puzzle:
1, 2, 6, 42, 1806, ?
Which wasn't too difficult. But it gave me an idea of a really complicated series which I've brought forth above.
Personally, I don't think anybody will solve this puzzle =D
Like last time, copy and paste the answer in the comments below using this code:
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TYPE YOUR ANSWER HERE
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~ James Kanjo
UPDATE: To check your answer, simply type it in the text box below, and press enter:
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The second one is easy.
First one, not so much, especially with only three numbers given.
This reminds me of a sequence I gave to a friend once. He never figured it out.
12, 36, 324, 2916, 52488, ?
Timothy Foster — My Site ~ My Blog
Timothy Foster  @tfAuroratide
Auroratide.com  Go here if you're nerdy like me
You nailed the first one. Funnily, I came up with the formula n(n+1), but I realised that this is just the expansion of n^{2}+n *facepalms*
The second answer was not the one that my sequence follows, but the method you used is not unlike the one my sequence follows. So the number you gave is not the next number in the series. Great try though!
λ James Kanjo
_{Blog  Wikidot Expert  λ and Proud Web Developer  HTML  CSS  JavaScript}
After nearly two years of contemplating this problem, I've arrived at the conclusion that three given numbers isn't enough to establish a unique pattern. The series could follow a number of patterns each with a different end result. For instance, I can give you {1,1,2}, and at first it looks like Fibonacci's sequence, so the next number should be {3}. But, I could say it follows a_{n} = (n1)!, in which case the next number is {6}. Or, it could even be:
(1)Which would yield {0}. I suppose this could be done with any number of givens, but once four or five numbers are given, it becomes more grounded that one unique pattern exists for that particular series.
I found a pattern in yours, which I reveal below:
Timothy Foster  @tfAuroratide
Auroratide.com  Go here if you're nerdy like me
Wow Timothy! That's impressive! But that's not the pattern :(
To add to the problem, I'm going to provide the fourth number, and you need to identify the next number in the series.
λ James Kanjo
_{Blog  Wikidot Expert  λ and Proud Web Developer  HTML  CSS  JavaScript}
Dang, adding that fourth number makes this seem much more complicated!
I'll be back in two years, and maybe I'll have a solution then ;)
Timothy Foster  @tfAuroratide
Auroratide.com  Go here if you're nerdy like me
I think I finally solved it. At least, it fits so well there is no way it could be accidental.
Timothy Foster  @tfAuroratide
Auroratide.com  Go here if you're nerdy like me
NO WAY!!! YOU GOT IT!!!
That's F*CKING AMAZING!!!
λ James Kanjo
_{Blog  Wikidot Expert  λ and Proud Web Developer  HTML  CSS  JavaScript}
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