What is the equation for this pattern? (x=0, 2, 4, 16, 256, 65536)

  • Thread starter Qwerty
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In summary, the conversation is about finding an equation for a specific pattern that involves squaring itself. The solution is found to be 2^(2^x), with x=0 being the first number in the pattern.
  • #1
Qwerty
I have the following pattern and i need an equation that if x = 1 then y will equal the fist number of it, if x = 2, then y will equal the 2nd number.

The pattern is:
2, 4, 16, 256, 65536

Hint: It keeps squaring itself.

Any ideas?
 
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  • #2
Try writing all of them as powers of 2:
21,22,24,28,216

It should be easy from there...
 
  • #3
Try writing all of them as powers of 2
But this doesn't help me much becasue it doesn't find my numbers i need in an equation...
 
  • #4
f(x) = 2^2x
 
Last edited:
  • #5
f(x) = 2^2x
I had already thought of this but it does not work...Here is teh pattern it gives you.
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
Originally posted by PrudensOptimus
f(x) = 2^2x

No, that gives 22, 24, 26, ...

I don't think we should be giving explicit formulas anyway, since it sounds like homework.
 
  • #7
Qwerty, take NateTG's suggestion and first find a formula for just the exponents 1, 2, 4, 8, 16, ... from that you can derive a formula for the actual sequence.
 
  • #8
it turns out to be quite a function with an even nicer derivative
 
  • #9
I don't think we should be giving explicit formulas anyway, since it sounds like homework.
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)
 
  • #10
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)
Originally posted by Qwerty
I have the following pattern and i need an equation that if x = 1 then y will equal the fist number of it, if x = 2, then y will equal the 2nd number.

The pattern is:
2, 4, 16, 256, 65536

Hint: It keeps squaring itself.

Any ideas?
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.
but if you put first x=0 then this is the equation you were looking for this pattern.
 
  • #11
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.
Yes this is an error in my first post, 0 is supposed to be the first number subsituted into the equation.
 

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