# Can someone help me with this sequence?1 4 1 16 1

1. Jul 30, 2010

### Alexx1

Can someone help me with this sequence?

1 4 1 16 1 36

If I only look at the odd numbers it's: n^2

But I don't know how they get to '1'..

2. Jul 30, 2010

### sshzp4

Re: sequence

My answer would be to construct a step function for the y power term in x^y. The step function is a continuous representation of the integral of kronecker/dirac delta. The condition of the step is 'evenness' or 'oddness'.

1 8 1 64 1 216..

then my solution would be
$$n^{g};$$
$$g=\frac{2}{1+ exp(P*(3*binx_n -1)};$$
where binx_n is the last (right-most) digit in the binary representation of n; and P is a very large number (say 1e9). So in your case, the series will be: (read from left to right.. solution from matlab for 100 term series)

1 4 1 16 1 36 1 64 1 100 1 144 1 196 1 256 1 324 1 400 1 484 1 576 1 676 1 784 1 900 1 1024 1 1156 1 1296 1 1444 1 1600 1 1764 1 1936 1 2116 1 2304 1 2500 1 2704 1 2916 1 3136 1 3364 1 3600 1 3844 1 4096 1 4356 1 4624 1 4900 1 5184 1 5476 1 5776 1 6084 1 6400 1 6724 1 7056 1 7396 1 7744 1 8100 1 8464 1 8836 1 9216 1 9604 1 10000
...

So the big question here becomes, is the representation actually continuous (for it to be a nice analytical solution)? The presence of the binary makes that question hard to answer. Obviously, using the binary is an overkill. But there's a ton of ways to flip the exponential's sign given a number is odd or even. Any answer anyone else comes up with (your instructor, maybe), will be reducible to the given form.

Delightful little question. Hope this shows you how to consider such problems.

Edit: Note-My expression is different from what you need! If you want to test your understanding, then try to express this series with a continuous, differentiable expression:
1 4 1 64 1 36 1 512 1 100 1 1728 1 …

Last edited: Jul 30, 2010
3. Jul 30, 2010

### HallsofIvy

Re: sequence

So you are starting with $a_0$? There is nothing at all wrong with:
$a_n= 1$ if n is even, $(n+1)^2$ if n is odd.

If you insist upon a single formula,
$$a_n= \frac{1- (-1)^n}{2}+ \frac{1+ (-1)^n}{2} (n+1)^2$$

When n is even $(-1)^n= 1$ so $\frac{1- (-1)^n}{2}= 0/2= 0$ and$\frac{1+ (-1)^n}{2}= 2/2= 1$. Then $a_n= 0+ (n+1)^2= (n+1)^2$.

When n is odd $(-1)^n= -1$ so $\frac{1- (-1)^n}{2}= 2/2= 1$ and $\frac{1- (-1)^n}{2}= 0$. Then $a_n= 1+ 0(n+1)^2= 1$.

Last edited by a moderator: Jul 31, 2010
4. Jul 30, 2010

### arildno

Re: sequence

There are many ways to do this, the most compact one is probably:

$$a_{n}=n^{(1+(-1)^{n})},n=1,2...$$

5. Jul 30, 2010

### Alexx1

Re: sequence

Thanks!

6. Jul 30, 2010

### Alexx1

Re: sequence

Thank you very much!

7. Jul 30, 2010

### Dickfore

Re: sequence

The even members are $(2 n)^{2}, n = 1, 2, \ldots$. The odd ones are always 1.

So, we can write:

$$a_{n} = \left\{\begin{array}{rc} n^{2}, & n \, \mathrm{even} \\ 1, & n \, \mathrm{odd} \end{array}\right.$$

You can write it in this form:

$$a_{n} = \frac{n^{2} + 1}{2} + (-1)^{n} \, \frac{n^{2} - 1}{2} = \frac{1 - (-1)^{n}}{2} + n^{2} \, \frac{1 + (-1)^{n}}{2}$$

8. Jul 30, 2010

### Staff: Mentor

Re: sequence

It looks to me like there's a different formula for the odd-index elements, with a2n+1 = 1, for n = 0, 1, 2, ...

9. Jul 30, 2010

### Staff: Mentor

Re: sequence

10. Jul 30, 2010

### arildno

Re: sequence

Mine's better!

11. Jul 30, 2010

### Mentallic

Re: sequence

I'm not exactly sure, but I can think of a way to get back to the number by.. say...

$$\frac{1+(-1)^{n-1}}{2}$$

In this formula, when n is odd, then we have 1, but when n is even we get 0.

So... we could use something like this

$$(S(n/2)-1).\frac{(1+(-1)^{n-1})}{2}+1$$

seems like it would be an appropriate formula for your problem, where S(n/2) is the sequence 4,16,36 for n=1,2,3.

So if you find a formula for the sequence S(n)=4,16,36 which I couldn't figure out (I'm pretty bad at sequences) then you can find a formula for the sequence you've shown by filling in S(n/2).

12. Jul 30, 2010

### Mentallic

Re: sequence

And mines probably the worst of all! :yuck:

But yes, I really like your formula.

13. Jul 30, 2010

### CRGreathouse

Re: sequence

Yeah, well, I beat you to it on the other thread. :tongue:

14. Jul 30, 2010

Re: sequence