What is the next number in the sequence?

joeyar

2, 8, 62, 622, 7772, ....

daskalou

116584?

Jimmy Snyder

I get

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

eom

daskalou

Ahh, good work, you're right.

joeyar

Yes, jimmysnyder got it. Well done mate.

RandallB

And quick too,
did you use any "special logic” to guide your judgment to a solution
or was it random attempts and personal "feel".

Jame

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 realised 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. :)

Borek

11111. These are roots of the following polynomial:

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



Borek
--
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Jame

Ah, how could I have missed something so obvious!

Jimmy Snyder

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.

Jimmy Snyder

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

Borek

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.

YesIam

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.