# Sequence: Periodic or Not?

1. Apr 12, 2007

### Asclepius

Let me preface this by saying this is not a homework problem or anything, although it may look like it to some. Also, I don't have much of a math background (Calc I & II, Linear Algebra), but I don't think this problem requires much knowledge of "higher," math; just some good problem solving skills. I'd be real greatful to anyone who could throw me some hints at where to go with this problem. Thanks a bunch in advance!

So anyway, here it is:

Consider sequence $$a_{n}=2^({2}^{n})$$. Let $$b_{n}$$ be the first digit of $$a_{n}$$. Determine whether the sequence $$b_{n}$$ is periodic.

I'm sure this is very elementary, but would appreciate all help/sympathy. :tongue2:

Last edited: Apr 12, 2007
2. Apr 12, 2007

### Asclepius

By the way, it's 2^2^n; Two raised to two, where the exponent "2" is raised to the n-th power.

3. Apr 13, 2007

### matt grime

$$2^{2^n}$$

tex is just like maths - brackets are important.

4. Apr 13, 2007

### tehno

Conversion into binary numeral system may help you to prove that bn can't be periodic.

5. Apr 15, 2007

### Asclepius

Thanks, tehno.

6. Apr 16, 2007

### robert Ihnot

techno: Conversion into binary numeral system may help you to prove that bn can't be periodic.

I wonder about that. What is being asked is The First Digit, and that first digit in the binary system is always periodic, since it must be "1."

7. Apr 16, 2007

### HallsofIvy

Are you assuming that "first digit" means leading digit? I would interpret it as "ones place digit".

8. Apr 16, 2007

### robert Ihnot

Halls of Ivy: Are you assuming that "first digit" means leading digit? I would interpret it as "ones place digit".

Something like that.
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9. Apr 16, 2007

### tehno

I understood what was being asked.
The last digit of $$2^{2^n}$$ is always 6 (easy to prove that).
In binary numeral system that means that the number can be always represented as "1...111".It can be shown,that any sequence formed of digits at any fixed place in between ,can't be periodic.And this is the stronger claim than OP's.The proof isn't short,though.