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- From: stevef@juliet.ll.mit.edu ( Steve Finch )
- Subject: A problem of Erdos
- Message-ID: <STEVEF.93Jan25170058@oswald.juliet.ll.mit.edu>
- Originator: dan@symcom.math.uiuc.edu
- Sender: Daniel Grayson <dan@math.uiuc.edu>
- X-Submissions-To: sci-math-research@uiuc.edu
- Organization: M.I.T. Lincoln Lab - Group 43
- X-Administrivia-To: sci-math-research-request@uiuc.edu
- Approved: Daniel Grayson <dan@math.uiuc.edu>
- Date: Mon, 25 Jan 1993 22:00:58 GMT
- Lines: 33
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- What is the status of the following problem due to Erdos? Thank you.
- Please respond via e-mail.
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- Denote by f(n) the largest integer so that, for every sequence of real
- numbers
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- a(1), a(2), ... , a(n)
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- with a(k) nonzero for all k, one can always select f(n) of them
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- a(j(1)), a(j(2)), ... , a(j(f(n)))
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- with the property that
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- a(j(p)) + a(j(q)) != a(j(r)) for all 1 <= p < q < r <= f(n).
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- PROVE THAT: f(n) = [ (n + 2)/2 ]
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- NOTE: != means "not equal to" and [x] means "greatest integer <= x".
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- REFERENCE:
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- P. Erdos, "Extremal Problems in Number Theory", Proc. Symposia in Pure Math.,
- Theory of Numbers, vol. 8, AMS, pp. 181 - 189.
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