- #1

joeyar

- 53

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter joeyar
- Start date

- #1

joeyar

- 53

- 0

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

- #2

daskalou

- 9

- 0

116584?

- #3

Jimmy Snyder

- 1,095

- 19

I get

117644

n^(n-1) - n + 2

eom

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

And quick too,I get

117644

n^(n-1) - n + 2

eom

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

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

[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

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.And quick too,

did you use any "special logic” to guide your judgment to a solution

or was it random attempts and personal "feel".

- #11

Jimmy Snyder

- 1,095

- 19

That's quite a coincidence. It turns out that 11111 is also the next number in the sequence:11111. These are roots of the following polynomial:

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

1 -19577 99504914 -60788218692 3929719423336 -34258540436320 53282917476608

- #12

Borek

Mentor

- 29,138

- 3,771

- #13

YesIam

- 42

- 0

Share:

- Last Post

- Replies
- 10

- Views
- 351

- Replies
- 37

- Views
- 2K

- Replies
- 25

- Views
- 558

- Last Post

- Replies
- 7

- Views
- 578

- Last Post

- Replies
- 2

- Views
- 395

- Last Post

- Replies
- 1

- Views
- 246

- Last Post

- Replies
- 12

- Views
- 288

- Replies
- 4

- Views
- 331

- Replies
- 18

- Views
- 410

- Replies
- 32

- Views
- 759