What is the next number in the sequence?

  • Thread starter joeyar
  • Start date
  • #1
joeyar
53
0
2, 8, 62, 622, 7772, ...
 

Answers and Replies

  • #2
daskalou
9
0
116584?
 
  • #3
Jimmy Snyder
1,095
19
I get

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

eom
 
  • #4
daskalou
9
0
I get

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

eom

Ahh, good work, you're right.
 
  • #5
joeyar
53
0
Yes, jimmysnyder got it. Well done mate.
 
  • #6
RandallB
1,550
0
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".
 
  • #7
Jame
42
0
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. :)
 
  • #8
Borek
Mentor
29,138
3,771
11111. These are roots of the following polynomial:

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

:tongue:
 
  • #9
Jame
42
0
Ah, how could I have missed something so obvious!
 
  • #10
Jimmy Snyder
1,095
19
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
Jimmy Snyder
1,095
19
11111. These are roots of the following polynomial:

[tex]f(x) = x^6-19577x^5+99504914x^4-60788218692x^3+3929719423336x^2-34258540436320x+53282917476608[/tex]
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
Borek
Mentor
29,138
3,771
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
YesIam
42
0
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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