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1 June 2002: Page created with first letter converted to PDF. Need to add text version.
1 June 2002 (evening): Added text version.
Paul Erdös
(19131996)
[New York Times obituary]
Correspondence
Here [PDF] is a two page letter
that Paul Erdös wrote to Klarner when Erdös visited the University
of Nebraska at Lincoln (UNL). At that time Klarner was on leave from
the University of Nebraska. The letter is dated 13 January 1992, and
is written on UNL math department stationery.
Transcription:
[letterhead]
University of Nebraska Lincoln
Department of Mathematics and Statistics
810 Oldfather Hall
Lincoln, Nebraska
Dear Klarner (1992 I 13)
I just arrived here, will be here this week and will preach tomorrow
and on thursday. I was sorry to hear that you are in Eindhoven, I was
there in 1967 with my mother. Please give my regards to my many friends
there. Just by accident I came across a few minutes ago an old paper of
mine (Extremal problems in number theory, 1965 Amer Math Soc meeting
in Pasadena 1963 Vol VIII), among others I consider there
the following problem:
Let a_{1} < a_{2} < ... < a_{n} be n integers
I prove that you can always find n/3 of them which are sum free ie
a_{i1}, ... , a_{ik}, k > n/3,
a_{ir1} +
a_{ir2} ≠
a_{ir3}, n/3 can probably be improved
but not beyond 11/28 n, then I also ask how many a's can you find
that the sum of two of them a_{i} + a_{j} ≠
a_{k}, i ≠ j, ie
a_{i1}, ... , a_{ik}, but the
sum of two a's is not one of the
a_{1}, ... ,a_{n} not only not one of the
a_{i1}, ... , a_{ik}.
i = j must be forbidden otherwise a_{i} = 2^{i} would
kill the problem. I state that you proved that k > c log n, probably
much more is true.
Let a_{1} < a_{2} < ... < a_{n} < n,
k > ε n is it true that if n > n_{0}(ε) then there are
three a's which have pairwise the same least common multiple? Is this
a good problem or trivially true or false??
Another such question
a_{1} < a_{2} < ... < a_{t} ≤ n, and no
a_{j} = a_{i} + a_{i+1} + ... + a_{i+r}
max t = ? I first thought that max t = n/2 + 1 but Pomerance found for
n=4m (m odd) 2m+2 such numbers m1, m, m+1, (3m1)/2, (3m+1)/2, 2m, ...
and all the integers 2m ≤ t ≤ which are not forbidden (exactly
4 are forbidden e.g. m=9 n=20 4 5 6 7 8 10 12 14 16 17 19 20.
15= 4+8 18=10+8 are forbidden. Can you get a good bound for t?
(n/2)(1+o(1)) is true?
I hope your health is good, you can always reach me at my Budapest adress or
c/o Dr R L Graham Bell Laboratories Murray Hill New Jersey 07974
Kind regards to all, au revoir
E. P.
You probably know that the Rados both died, but their son Peter + family
lives in the house.
