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A058241
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Number of sets of n positive integers that can be placed along a circle such that the set of sums of adjacent integers forms { 1, 2, ..., n^2-n+1 }.
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4
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1, 1, 1, 2, 1, 5, 0, 6, 4, 6, 0, 18, 0, 20, 0, 0, 6, 51, 0, 42, 0, 0, 0
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OFFSET
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1,4
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COMMENTS
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a(1)=1, a(2)=1.
Conjecture: for n > 2, p prime, e > 0, if n-1 is of the form p^e then a(n) > 0, otherwise a(n)=0.
From _Zhao Hui Du_, Mar 18 2019: (Start)
Conjecture: for n > 2, p prime, e > 0, if n-1 is of the form p^e then a(n) = A000010(n^2-n+1)/(6e), otherwise a(n)=0.
If a(n) is nonzero, a finite projective plane of order n-1 could be constructed.
Brute-force enumeration shows a(21)=0.
The Bruck-Ryser Theorem shows that if a finite projective plane of order q exists and q is congruent to 1 or 2 (mod 4), then q must be the sum of two squares. We could get a(22)=a(23)=0 from that theorem.
(End)
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LINKS
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Leonard E. Dickson, Problem 142, The American Mathematical Monthly, Vol. 14, No. 5 (May, 1907), pp. 107-108.
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EXAMPLE
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For n=3, we can choose a set { 1, 2, 4 } and place them along a circle as (1,4,2). Then the sums of adjacent numbers give all numbers from 1 to 7=3*(3-1)+1: { 1=1, 2=2, 3=1+2, 4=4, 5=1+4, 6=2+4, 7=1+2+4 }. Since such set is unique, a(3) = 1.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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_Naohiro Nomoto_, Jan 16 2001
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EXTENSIONS
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More terms from Rustem Aidagulov (rustem53(AT)mail.ru), Sep 06 2005 and Jan 01 2006
a(21)-a(23) from _Zhao Hui Du_, Mar 17 2019
Edited by _Max Alekseyev_, Jul 23 2019
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STATUS
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approved
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