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Re: Can you give a proof?

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  • video_ranger
    ... n)) ... Yes, writing n as a three digit binary number n = b1b2b3: (a,b,c)*n = (a-b1,b-b2,c-b3) and the sum over all three digit binary numbers can be
    Message 1 of 2 , Dec 2, 2008
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      --- In mathforfun@yahoogroups.com, "ramsey2879" <ramsey2879@...>
      wrote:
      >
      > A.Let n go from 0 to 7
      > B. Using the 3-bit binary values for 0 to 7, [000,001 ... 110,111]
      > let (a,b,c)*n = (d,e,f) =(a-(4 and n)/4},b-(2 and n)/2,(c-(1 and
      n))
      >
      > i.e.
      > (2,3,4)*0 = (2,3,4)
      > (2,3,4)*1 = (2,3,3)
      > (2,3,4)*2 = (2,2,4)
      > (2,3,4)*3 = (2,2,3)
      > ...
      > (2,3,4)*6 = (1,2,4)
      > (2,3,4)*7 = (1,2,3)
      >
      > Let F(a,b,c) = T(a)*T(b)*T(c) where T(x) = x(x+1)/2 are the ordinary
      > triangular numbers
      >
      > Prove that Sum (n = 0 to 7) F((a,b,c)*n) = (abc)^2


      Yes, writing n as a three digit binary number n = b1b2b3:

      (a,b,c)*n = (a-b1,b-b2,c-b3)

      and the sum over all three digit binary numbers can be written as the
      nested sum:

      Sum (b1=0,1) Sum(b2=0,1) Sum(b3=0,1) F(a-b1,b-b2,c-b3)
      =
      Sum (b1=0,1) Sum(b2=0,1) Sum(b3=0,1) T(a-b1)T(b-b2)T(c-b3)
      =
      (T(a)+T(a-1))*(T(b)+T(b-1))*(T(c)+T(c-1))

      but:

      T(a)+T(a-1) = (a/2)(a+1 + a-1) = a^2 etc.. QED.
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