A problem of Erdos

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1 total messages Started by Steve Finch Mon, 25 Jan 1993 22:00

#190

Author: Steve Finch
Date: Mon, 25 Jan 1993 22:00

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What is the status of the following problem due to Erdos?  Thank you.
Please respond via e-mail.


Denote by f(n) the largest integer so that, for every sequence of real
numbers

		a(1), a(2), ... , a(n)

with a(k) nonzero for all k, one can always select f(n) of them

		a(j(1)), a(j(2)), ... , a(j(f(n)))

with the property that

	a(j(p)) + a(j(q)) != a(j(r))   for all  1 <= p < q < r <= f(n).

PROVE THAT:  f(n) = [ (n + 2)/2 ]


NOTE:  != means "not equal to" and [x] means "greatest integer <= x".


REFERENCE:

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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