A058241 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 }.

1, 1, 1, 2, 1, 5, 0, 6, 4, 6, 0, 18, 0, 20, 0, 0, 6, 51, 0, 42, 0, 0, 0

Offset: 1

Author

Naohiro Nomoto, Jan 16 2001

Keywords

• nonn
• more

Comments

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)

Examples

			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.
		

Links

• Leonard E. Dickson, Problem 142, The American Mathematical Monthly, Vol. 14, No. 5 (May, 1907), pp. 107-108.
• D. Mehendale, Finite Projective Planes, arXiv:math/0611492 [math.GM], 2006-2015.
• Eric Weisstein's World of Mathematics, Perfect Difference Set
• Wikipedia, Bruck-Ryser theorem

Crossrefs

Cf. A000010.

Extensions

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