Looking for insights about this sequence:

[realitybugll]

1 2 3 4 5 6
1, 1, 2, 3/2, 4/(3/2), 5/(4/(3/2)), ...

I suppose there are a lot of variations, but the general idea is the terms are defined by:

f(n)*f(n+1)=n, where n is an integer. The top row is the term # (n), and the bottom one is the actual value of the term

Particularly, I am looking for a way to find the partial sums of the terms.Any responses are appreciated. Sorry for the unclear formatting...

 

[sylas]

1 2 3 4 5 6
1, 1, 2, 3/2, 4/(3/2), 5/(4/(3/2)), ...

I suppose there are a lot of variations, but the general idea is the terms are defined by:

f(n)*f(n+1)=n, where n is an integer. The top row is the term # (n), and the bottom one is the actual value of the term

Particularly, I am looking for a way to find the partial sums of the terms.Any responses are appreciated. Sorry for the unclear formatting...

Here's a start; closed forms for the odd and even terms. However, I have started your series counting from 0. That is, where you write f(1) I have f(0). Otherwise it's the same.

\begin{align*}<br /> f(2n) &amp; = 4^n n!^2 / (2n)! \\<br /> f(2n+1) &amp; = (2n+1)! / n!^2 / 4^n<br /> \end{align*}​

Try it.

Cheers -- sylas

 

[realitybugll]

sylas,

Wow! thank you.

Could you maybe give me a hint as to how you found that?

 

[sylas]

sylas,

Wow! thank you.

Could you maybe give me a hint as to how you found that?

Sure. Basically, I noted that the form of your fractions is as follows.

\frac{n(n-2)(n-4)(n-6)...}{(n-1)(n-3)(n-5)(n-7)...}​

To make life easy for myself, I started by looking only at even values of n. What I actually did was consider cases for 2n, so the equation becomes

\frac{2n(2n-2)(2n-4)(2n-6)...2}{(2n-1)(2n-3)(2n-5)(2n-7)...1}​

This looks a lot like a factorial, so I immediate thought about dividing everything by 2. Once I have a formula for the top line, I can see I will be able to get the bottom line by dividing (2n)! by the top line, so I just focus on the top line. There are n terms being multiplied, so the top line is

2^n n(n-1)(n-2)(n-3)...1 = 2^n n!​

The bottom line is therefore

\frac{(2n)!}{2^n n!}​

So we divide these two, and obtain:

\frac{(2^n n!)^2}{(2n)!} = \frac{4^n n!^2}{(2n)!}​

Given the way you numbered equations, this would actually be term number 2n+1. So I simply started numbering from 0. I'm also a pure mathematician, and for programming I like C better than Fortran... so I usually start counting from zero anyway. It often simplifies a problem like this.

Finding the equation for the odd terms was a breeze, using f(n)*f(n-1) = n, which you was your defining relation (adjusted to start counting at zero).

Cheers -- sylas

 

[realitybugll]

Thanks, I read over it and then went through it myself and got the same thing