- #1

The pattern is:

2, 4, 16, 256, 65536

Hint: It keeps squaring itself.

Any ideas?

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- Thread starter Qwerty
- Start date

- #1

The pattern is:

2, 4, 16, 256, 65536

Hint: It keeps squaring itself.

Any ideas?

- #2

NateTG

Science Advisor

Homework Helper

- 2,452

- 7

2

It should be easy from there...

- #3

But this doesn't help me much becasue it doesn't find my numbers i need in an equation...Try writing all of them as powers of 2

- #4

PrudensOptimus

- 641

- 0

f(x) = 2^2x

Last edited:

- #5

I had already thought of this but it does not work...Here is teh pattern it gives you.f(x) = 2^2x

4,16,64,256

This is the pattern I want...

2, 4, 16, 256, 65536

There is not supposed to be 64 and it also skips 2

- #6

Ambitwistor

- 841

- 1

Originally posted by PrudensOptimus

f(x) = 2^2x

No, that gives 2

I don't think we should be giving explicit formulas anyway, since it sounds like homework.

- #7

Ambitwistor

- 841

- 1

- #8

lastlaugh

- 19

- 0

it turns out to be quite a function with an even nicer derivative

- #9

Acually no, it's not for homework....its for a project I am working on(programming).I don't think we should be giving explicit formulas anyway, since it sounds like homework.

And I don't need your help anymore, I found the answer to my question myself.

Answer: 2^(2^x)

- #10

MathematicalPhysicist

Gold Member

- 4,699

- 369

Originally posted by Qwerty

Acually no, it's not for homework....its for a project I am working on(programming).

And I don't need your help anymore, I found the answer to my question myself.

Answer: 2^(2^x)

if you first put x=1 (as you requested) to this equation you get 2^(2^1)=4 which is not the first number 2 ofcourse.Originally posted by Qwerty

The pattern is:

2, 4, 16, 256, 65536

Hint: It keeps squaring itself.

Any ideas?

but if you put first x=0 then this is the equation you were looking for this pattern.

- #11

Yes this is an error in my first post, 0 is supposed to be the first number subsituted into the equation.if you first put x=1 (as you requested) to this equation you get 2^(2^1)=4 which is not the first number 2 ofcourse.

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