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Next Number?

  1. Oct 17, 2005 #1
    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. jcsd
  3. Oct 18, 2005 #2
    i guess the next number is... 165
     
  4. Oct 18, 2005 #3
    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, ...
     
  5. Oct 18, 2005 #4
    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
  6. Oct 18, 2005 #5
    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
     
  7. Oct 18, 2005 #6
    It is a typo, it works when n is a perfect square.
     
  8. Oct 18, 2005 #7
    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.
     
  9. Oct 18, 2005 #8
    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
  10. Oct 19, 2005 #9
    I still don't get it. In your sequence, T(4) is 90, not 15.
     
  11. Oct 19, 2005 #10
    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,...
     
  12. Oct 19, 2005 #11
    ... or it can be this way also...

    nth term = (2*n^3 - 3*n^2 + n)/2 , where n=1,2,3,4,5,... :cool:
     
  13. Oct 20, 2005 #12
    It works if you do it for n= 0^2, 1^2, 2^2, ...

    then 2^2 = 15, but 4^2 = 90.
     
  14. Oct 20, 2005 #13
    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...
     
  15. Oct 20, 2005 #14
    Work it out for me for 3^2 and maybe I'll finally get it.
     
  16. Oct 20, 2005 #15
    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.
     
  17. Oct 20, 2005 #16
    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.
     
  18. Oct 20, 2005 #17
    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.
     
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