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Math Help - Challenge

  1. #1
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    Question Challenge

    Consider a function that behaves like this:
    F[0,y]={1,1,1,1,1..}
    F[1,y]={1,2,3,4,5..}
    F[2,y]={1,3,6,10,15..}
    F[3,y]={1,4,10,20,35..}
    F[4,y]={1,5,15,35,70..}
    ...
    which is simply F[x,y]=Binomial[x+y,x].

    Now the challenge is to come up with an equation that explains this behavior:
    G[1,y]={0,1,2,3,4,5....}
    G[2,y]={0,1,1,3,3,3,6,6,6,6,10,10,10,10,10,15,15,15,15,1 5,15,21...}
    G[3,y]={0,1,1,1,4,4,4,4,4,4,10,10,10,10,10,10,10,10,10,1 0,20,20....}
    G[4,y]={0,1,1,1,1,5,5,5,5,5,5,5,5,5,5,15...}
    ...
    It shouldn't be too hard to figure out the pattern, but coming up with a function is the tricky part I can't figure out. Anybody want to take a shot at it? This isn't homework.
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  2. #2
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    Re: Challenge

    Can you invert figurate numbers, in general?
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  3. #3
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    Re: Challenge

    Quote Originally Posted by a tutor View Post
    Can you invert figurate numbers, in general?
    I know what figurate numbers are, but I don't understand the question.
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  4. #4
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    Re: Challenge

    G[3,4] = 4 = F[3,1]
    G[3,10] = 10 = F[3,2]
    G[3,20] = 20 = F[3,3]

    If you solve

    n(n+1)(n+2)/6 =4

    n(n+1)(n+2)/6 =10

    n(n+1)(n+2)/6 =20

    you get n=2,n=3 and n=4.
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  5. #5
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    Re: Challenge

    Quote Originally Posted by a tutor View Post
    G[3,4] = 4 = F[3,1]
    G[3,10] = 10 = F[3,2]
    G[3,20] = 20 = F[3,3]

    If you solve

    n(n+1)(n+2)/6 =4

    n(n+1)(n+2)/6 =10

    n(n+1)(n+2)/6 =20

    you get n=2,n=3 and n=4.
    Not sure I follow. G[3,4] should equal 4 as well as G[3,5]=4, G[3,6]=4, G[3,7]=4, G[3,8] and G[3,9]=4. G[3,10] should then equal 10, G[3,11]=10, etc.
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  6. #6
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    Re: Challenge

    Yes but if you solve

    n(n+1)(n+2)/6=4
    n(n+1)(n+2)/6=5
    n(n+1)(n+2)/6=6
    n(n+1)(n+2)/6=7
    n(n+1)(n+2)/6=8
    n(n+1)(n+2)/6=9

    and round down you get n=2 in every case.
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  7. #7
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    Re: Challenge

    Quote Originally Posted by a tutor View Post
    Yes but if you solve

    n(n+1)(n+2)/6=4
    n(n+1)(n+2)/6=5
    n(n+1)(n+2)/6=6
    n(n+1)(n+2)/6=7
    n(n+1)(n+2)/6=8
    n(n+1)(n+2)/6=9

    and round down you get n=2 in every case.
    Ah, I understand. I haven't tried it but I assume the continuation for other values of x in G[x,y] would be:

    x=4: n(n+1)(n+2)(n+3)/24
    x=5: n(n+1)(n+2)(n+3)(n+4)/120
    ...
    But then there's the problem of isolating an equation for n for any x. The higher x is, the more difficult it is to solve for n. Is this correct?
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