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- Newsgroups: sci.math
- Path: sparky!uunet!munnari.oz.au!manuel!gerry@macadam.mpce.mq.edu.au
- From: gerry@macadam.mpce.mq.edu.au (Gerry Myerson)
- Subject: Re: Partition number question
- Message-ID: <1992Jul23.235803.5110@newshost.anu.edu.au>
- Sender: news@newshost.anu.edu.au
- Organization: ceNTRe for Number Theory Research
- References: <4574@balrog.ctron.com>
- Date: Thu, 23 Jul 92 23:58:03 GMT
- Lines: 28
-
- In article <4574@balrog.ctron.com>, wilson@web.ctron.com writes:
- >
- >
- > For n >= 0, let p(n) be the unrestricted partition number of n.
- >
- > 1. For all k >= 0, is it true that
- >
- > a. 5 | p(5k+4)
- > b. 7 | p(7k+5)
- > c. 11 | p(11k+6)
-
- Yes. These are due to Ramanujan.
-
- > 2. Given (1) is true, are there other triples (a > 0, b > 0, c >= 0)
- > such that a | p(bk+c) for all k >= 0?
-
- There are some with higher powers of 5, 7, and 11, e.g., 25 divides p(25k+24).
- There may be others--I'm no expert here. Latest edition of Hardy and Wright,
- The Theory of Numbers, should be a good place to start.
-
- > 3. Why does 11 divide so many small partition numbers?
- > --
- > David W. Wilson (wilson@ctron.com)
- >
- > Disclaimer: "Truth is just truth...You can't have opinions about truth."
- > - Peter Schikele, introduction to P.D.Q. Bach's oratorio "The Seasonings."
-
- Gerry Myerson
-