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- Newsgroups: sci.math
- Path: sparky!uunet!newsgate.watson.ibm.com!yktnews!admin!yktnews!victor
- From: victor@watson.ibm.com (Victor Miller)
- Subject: Re: An Interesting Function
- Sender: news@watson.ibm.com (NNTP News Poster)
- Message-ID: <VICTOR.92Nov12104834@terse.watson.ibm.com>
- In-Reply-To: sasrdt@shewhart.unx.sas.com's message of Wed, 11 Nov 1992 18:37:05 GMT
- Date: Thu, 12 Nov 1992 15:48:34 GMT
- Reply-To: victor@watson.ibm.com
- Disclaimer: This posting represents the poster's views, not necessarily those of IBM
- References: <1992Oct29.001255.35683@Cookie.secapl.com> <BxKEDt.sL@unx.sas.com>
- Nntp-Posting-Host: terse.watson.ibm.com
- Organization: IBM, T.J. Watson Research Center
- Lines: 48
-
- >>>>> On Wed, 11 Nov 1992 18:37:05 GMT, sasrdt@shewhart.unx.sas.com (Randall D. Tobias) said:
- Randall> Originator: sasrdt@shewhart.unx.sas.com
- Randall> Nntp-Posting-Host: shewhart.unx.sas.com
-
-
- Randall> In article <1992Oct29.001255.35683@Cookie.secapl.com>, frank@Cookie.secapl.com (Frank Adams) writes:
- Frank> Consider the function W(n) = sum(0<i<n,mod(n,i)). The problem is to find an
- Frank> asymptotic expression for W(n).
- Frank>
- Frank> It is pretty clear that W(n) = O(n^2). In fact, we can write
- Frank> W(n)=c n^2 + o(n^2). The question is, what is c? I know the answer, and
- Frank> will post it later if no one gets it.
- Frank>
- Frank> I think in fact W(n) = c n^2 + O(n), but I can't prove it. It seems that
- Frank> W(n) /= c n^2 + d n + o(n); the O(n) stuff "wanders around".
-
- Randall> Well? I got as far as empirically verifying the
-
- Randall> W(n) /= c n^2 + d n + o(n)
-
- Randall> for n up to 100000 or so. But i haven't got the number theory know-how
- Randall> to go much further. Pretty function, though.
-
- Randall> --
-
- Randall> Randy Tobias SAS Institute Inc. sasrdt@unx.sas.com
- Randall> (919) 677-8000 x7933 SAS Campus Dr. 72450.2545@compuserve.com
- Randall> (919) 677-8123 (Fax) Cary, NC 27512-8000 norat@aol.com
-
- Randall> ... just my $(-exp(2*sqrt(-1)*arcos(0))/(((2**(2 + 1)) - 1)**2 + 1)).
-
- The answer is W(n) = (1 - 1/12 \pi^2) n^2 + O(n log n). This may be
- seen as follows:
-
- W(n) = \sum_{i=1}^n (n - i [n/i]) = n^2 - \sum_{i=1}^n i [n/i].
-
- The latter sum is = \sum_{i=1}^n \sum_{k=1, i|k}^n i =
- \sum_{k=1}^n\sum_{i|k} i = \sum_{k=1}^n \sigma(k), where \sigma(k) is
- the sum of the divisors of k. By Hardy and Wright Theorem 234 (page
- 266 in the Fourth edition), the latter sum is = (1/12) \pi^2 + O(n log n).
-
-
- --
- Victor S. Miller
- Bitnet: VICTOR at WATSON
- Internet: victor@watson.ibm.com
- IBM, TJ Watson Research Center
- "Great artists steal; lesser artists borrow" Igor Stravinsky
-