[seqfan] Re: identity

[Richard Mathar]

On the observation
sum( phi(2*k+1)*round((n-1)/(2*(2*k+1))), k = 0..(n/2)-1 ) = n^2/4
in http://list.seqfan.eu/pipermail/seqfan/2010-February/003583.html 
(and its obvious generalization to odd n if the r.h.s is replaced by [n/2]^2 ):


This looks vaguely like an appication of (1) in Mercier's article
"Sum containing the fractional parts of numbers"
Rocy Mountain J. of Math, 15 (2) (1985) 513
http://projecteuclid.org/euclid.rmjm/1250127225
http://dx.doi.org/10.1216/RMJ-1985-15-2-513
MR823262

sum_{k=0.. n/2-1} phi(2k+1)* [ (n+2k-2)/(2(2k+1))]
=
sum_{k=1,3,5,7.. n-1} phi(k)* [ (n-3+k)/(2k)]
appears to be of the form on his right hand side.

Another format of proof might observe that the sum is underneath a parabolic
section of a curve in the (k,n) coordinates. So writing phi(k) by mobieus
inversion might allow resummation over the divisors of n..
(see P Haukkanen in Int J Math Math Sci 19 (2) (1996) 209-218 ,
http://www.doaj.org/doaj?func=abstract&id=292470
http://dx.doi.org/10.1155/S0161171296000312
 for a useful collection of transformations..)

Richard Mathar