General apologies to everybody --- the "Send" and "Back to Inbox" buttons are a bit too close for comfort on my mailer window, so I occasionally hit the wrong one. Apparently several people have met their partners on dating sites as a result of similar confusion --- can't see that happening to me here, though! WFL On 2/18/12, David Wilson <davidwwilson@comcast.net> wrote:
Fred:
Did you add anything here? Asking only because I have come to expect gems whenever you chime in.
On 2/18/2012 4:33 PM, Fred lunnon wrote:
On 2/18/12, Michael Reid<reid@gauss.math.ucf.edu> wrote:
Letting mu be the Moebius function, and nu be the Mertens function defined as
nu(n) = SUM(k = 1..n, mu(k))
Then I notice
[1] SUM(k = 1..n, mu(k)*[n/k]) = 1
[2] SUM(k = 1..n, nu([n/k])) = 1
where [x] is the floor of x. I wondered if there was any simple relationship linking [1] and [2], or perhaps some relationship to the Sure, write the second sum as a double sum, using the definition of nu(-) , and then switch the order of summation.
Moebius transform
a(n) = SUM(d|n; b(d))<=> b(n) = SUM(d|n; mu(n/d)*a(d)).
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