# Next Number?

1. Oct 17, 2005

### nnnnnnnn

0, 3, 15, 42, 90, ...

What number comes next?

There's something interesting about these numbers (at least to me). <- thats not a clue its just that this isnt just some random series, theres more to it that is mathematical, not personal or a bus route...

Last edited: Oct 17, 2005
2. Oct 18, 2005

### croxbearer

i guess the next number is... 165

3. Oct 18, 2005

### Jimmy Snyder

I get 165 too. The formula is (2 * n + 1) * T(n), where T(n) is the sum of the first n digits, n = 0, 1, 2, ...

4. Oct 18, 2005

### nnnnnnnn

That works too, the formula I was using is T(0) = 0, T(n) = (T(n-1))/3 + n) * 3. The interesting part was that if you add consecutive numbers, they add to the next set of consecutive numbers, starting with a perfect square and going to the number before the next perfect square. Like 1+2 = 3 and 4+5+6 = 7+8 and so on in a pyramid sort of way.

Last edited: Oct 18, 2005
5. Oct 18, 2005

### Jimmy Snyder

Is there a typo in this? It doesn't seem to work. For instance

T(1) = (T(0)/3) + 1) * 3 = (0/3 + 1) * 3 = 1 * 3 = 3 is OK, but
T(2) = (T(1)/3) + 2) * 3 = (3/3 + 2) * 3 = 3 * 3 = 9

6. Oct 18, 2005

### nnnnnnnn

It is a typo, it works when n is a perfect square.

7. Oct 18, 2005

### Jimmy Snyder

You lost me there. Can you show me an example of it working? It seems to get the wrong answer for all n greater than 1, including perfect squares.

8. Oct 18, 2005

### nnnnnnnn

I miss stated it, it's: T(0) = 0, T(n) = (T( (n^(1/2) -1 )^2 )/3 + n) * 3, when n^(1/2) is an integer.

so for n = 4, its T(4) = ( T(1)/3 + 4 )*3 = (3/3 + 4)*3 = 15

It can also be n + (n + 1) + ... + (n + n^(1/2)) = (n + n^(1/2) + 1) + ... + ( n + 2*n^(1/2) )

Last edited: Oct 18, 2005
9. Oct 19, 2005

### Jimmy Snyder

I still don't get it. In your sequence, T(4) is 90, not 15.

10. Oct 19, 2005

### croxbearer

i think it would be simpler to express the formula like this:

nth term = (2*n^3 + 3*n^2 + n)/2 , where n=0,1,2,3,4,...

11. Oct 19, 2005

### croxbearer

... or it can be this way also...

nth term = (2*n^3 - 3*n^2 + n)/2 , where n=1,2,3,4,5,...

12. Oct 20, 2005

### nnnnnnnn

It works if you do it for n= 0^2, 1^2, 2^2, ...

then 2^2 = 15, but 4^2 = 90.

13. Oct 20, 2005

### nnnnnnnn

I think that would make it simpler too. Before I figured out the recursive function, I was using excel with the formula n + (n+1) + ... + (n+n^(1/2)), that was a pain because you can't just copy and paste...

14. Oct 20, 2005

### Jimmy Snyder

Work it out for me for 3^2 and maybe I'll finally get it.

15. Oct 20, 2005

### nnnnnnnn

T(n) = (T( (n^(1/2) -1 )^2 )/3 + n) * 3

T(0) = 0
T(1) = 3
T(4) = 15

T(3^2) = ( T(2^2) )/3 + 3^2) * 3 = (15/3 + 9) * 3 = 42

I think the problem was that I knew what I was thinking but then I didn't clearly state it. I was intending that T(n-1) would be the next lowest perfect square, so if n=9, I had n-1=4, even though I didnt state that until later. The most recent post should be an accurate formula.

16. Oct 20, 2005

### Jimmy Snyder

Thanks nnnnnnnn, I finally get it. You could simplify your own formula greatly as follows:

G(0) = 0
G(n) = G(n-1) + 3n^2, n = 1, 2, 3, ...

Essentially this is your formula with the factor of 3 distributed and n replaced with its square root.

For example:
G(0) = 0
G(1) = G(0) + 3 * 1 * 1 = 0 + 3 = 3
G(2) = G(1) + 3 * 2 * 2 = 3 + 12 = 15
G(3) = G(2) + 3 * 3 * 3 = 15 * 27 = 42
G(4) = G(3) + 3 * 4 * 4 = 42 + 48 = 90
etc.

17. Oct 20, 2005

### nnnnnnnn

I saw this too, and it would have been much easier to use. When I noticed the pattern (1+2=3, 4+5+6=7+8...), the perfect squares stuck out and then when I came up with the recursive function I was looking at it more from a point of view of the squares than I was from the pov of a sequense. Anyways I decided to keep it that way (even though it is more confusing and harder for me to state what I meant) because it seemed more related to the pattern.