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Floor identity

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  • Sebastian Martin
    Hello all: I have found this identity with floor function: 2Floor[(n-1)/2]+Sum[Floor[n/i]*Floor[n/(n-i)],{i,1,n-1}]= Sum[Floor[n/i]+Floor[n/(n-i)],{i,1,n-1}]
    Message 1 of 1 , Feb 17, 2007
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      Hello all:

      I have found this identity with floor function:

      2Floor[(n-1)/2]+Sum[Floor[n/i]*Floor[n/(n-i)],{i,1,n-1}]=
      Sum[Floor[n/i]+Floor[n/(n-i)],{i,1,n-1}]

      Is trivial or have any interest?

      Sincerely

      Sebastian Martin Ruiz


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