ruskey at cs.uvic.ca
Thu Feb 4 04:19:19 CET 2010
Does anyone know why I might find a proof of this or similar
identities? n is an even integer. In maple, so phi = totient
sum( phi(2*k+1)*round((n-1)/(2*(2*k+1))), k = 0..(n/2)-1 ) = n^2/4
The round is just floor((n+2k-2)/(4k+2)) in disguise. I could
probably use the fact that floor(a/b)-floor((a-1)/b) = 1 if b|a and
0 otherwise, but if it has already been done ...
Frank Ruskey e-mail: (last_name)(AT)cs(DOT)uvic(DOT)ca
Dept. of Computer Science office: 250-472-5794
University of Victoria fax: 250-472-5708
Victoria, B.C. V8W 3P6 CANADA WWW: http://www.cs.uvic.ca/~(last_name)
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