What is the next number in the sequence?

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The discussion revolves around identifying the next number in a sequence: 2, 8, 62, 622, 7772, with the proposed solution being 117644. The formula derived is n^(n-1) - n + 2, which fits the pattern of the sequence. Participants share their thought processes, noting the exponential growth and the increasing quotient between terms. One contributor highlights the importance of recognizing that the terms are close to small powers of integers, adjusted by small integers. Additionally, a polynomial is mentioned, with roots related to the sequence, and a humorous reference to the number 11111 being the next in the sequence is made. The conversation touches on the idea that any number could potentially fit into such sequences, with a nod to the common belief that 17 is often proposed as a next number in various sequences.
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2, 8, 62, 622, 7772, ...
 
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116584?
 
I get

117644
n^(n-1) - n + 2

eom
 
jimmysnyder said:
I get

117644
n^(n-1) - n + 2

eom

Ahh, good work, you're right.
 
Yes, jimmysnyder got it. Well done mate.
 
jimmysnyder said:
I get

117644
n^(n-1) - n + 2

eom
And quick too,
did you use any "special logic” to guide your judgment to a solution
or was it random attempts and personal "feel".
 
Got it too, love these brain teasers!

I first noticed there was exponential growth involved, I tried dividing the terms and noticed that the quotient of a term and its predecessor was increasing. I then did som algebra and noticed that expressions of the form n^A has a quotient approaching 1 as n approaches infinity, which doesn't fit this case. I then tried n^n and found that it met the increasing-quotient criteria, but the actual numbers for the cases of n = 1, 2, 3, 4 .. were a bit off. I then realized that it had to be n^(n-1) which gave me an almost perfect fit, except for a linearly increasing difference. This last term turned out to be (-n + 2). The next number therefore has to be n^(n-1) - n + 2 = 7^6 - 7 + 2 = 117644

When I do these kinds of puzzles I like to forget my knowledge of calculus and series and just do it the way I did when I was smaller and there was an exciting number-quiz in the newpaper. :)
 
11111. These are roots of the following polynomial:

f(x) = x^6-19577x^5+99504914x^4-60788218692x^3+3929719423336x^2-34258540436320x+53282917476608

:-p
 
Ah, how could I have missed something so obvious!
 
  • #10
RandallB said:
And quick too,
did you use any "special logic” to guide your judgment to a solution
or was it random attempts and personal "feel".
The fact that 8, 62, and 622 are all close to small powers of small integers, and off by 1, 2, and 3 was the key for me.
 
  • #11
Borek said:
11111. These are roots of the following polynomial:

f(x) = x^6-19577x^5+99504914x^4-60788218692x^3+3929719423336x^2-34258540436320x+53282917476608
That's quite a coincidence. It turns out that 11111 is also the next number in the sequence:

1 -19577 99504914 -60788218692 3929719423336 -34258540436320 53282917476608
 
  • #12
TBH that's not my idea. I believe originally it was claimed that 17 is the next number in every sequence, but I don't remember who was the author.
 
  • #13
Hi Borek and others - I just happened on this forum a couple of days ago. Maybe you all are way ahead of me...or maybe not. I thought it was common knowledge that any number can be a correct number in a series sequence like these. Almost like Borek says, except, "...17 can be...", rather than, "...17 is..." I think that can be chiseled in stone.
 
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