paul erdös

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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
(1913-1996)
[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 a₁ < a₂ < ... < a_n be n integers I prove that you can always find n/3 of them which are sum free ie a_i₁, ... , a_{i_k}, k > n/3, a_{i_r₁} + a_{i_r₂} ≠ a_{i_r₃}, 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_{i_k}, but the sum of two a's is not one of the a₁, ... ,a_n not only not one of the a_i₁, ... , a_{i_k}. i = j must be forbidden otherwise a_i = 2ⁱ would kill the problem. I state that you proved that k > c log n, probably much more is true.

Let a₁ < a₂ < ... < a_n < n, k > ε n is it true that if n > n₀(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 a₁ < a₂ < ... < 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 m-1, m, m+1, (3m-1)/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.