3,1

In a problem in the "Bundeswettbewerb 2002" competition there are 12 sticks of lengths 1,..,12 put in a ring in random order. It has to be proved that there are at least 3 consecutive sticks with total length not less than 20. A closer look shows that the total length is at least a(12)=21. The problem of the contest is a consequence of the following observation: every term a(n) is at least ceil(3*(n+1)/2), since n*a(n) >= sum{i=1..n}(p(i-1)+p(i)+p(i+1)) = 3*sum{i=1..n}(i) =3*n*(n+1)/2. So in the case n=12 we have (total length) >= a(12)=21 >= 20.

Thread "Zahlenkreis" in de.sci.mathematik, December 2001

T. J. Rolfe, P. W. Purdom, "An Alternative Problem for Backtracking and Bounding", Bulletin of the ACM SIG on Computer Science Education, Vol. 36, No. 4 (December 2004), pp. 83-84 [From Milan Stefanovic (stefke381(AT)gmail.com), Nov 19 2008]

Bundeswettbewerb Mathematik 2002 (1.4)

Let p be a permutation of 1..n and let g(p) be the maximum of the consecutive triple sums p(i-1)+p(i)+p(i+1), where p(0)=p(n) and p(n+1)=p(1). a(n) is the minimum of all the g(p) taken over all permutations p.

a(6)=11 because cycle 1-4-5-2-3-6- has sums 11,10,11,10,11,10 with max=11.

This example by Helmut Richter shows that a(14) = 24 is very likely: p = (1-8-11-4-9-10-2-12-5-6-13-3-7-14-) with g(p) = 11+4+9 = 24 as maximal three-sum.

Sequence in context: A140052 A070598 A124257 this_sequence A103092 A104523 A091886

Adjacent sequences: A066382 A066383 A066384 this_sequence A066386 A066387 A066388

nice,nonn

Rainer Rosenthal (r.rosenthal(AT)web.de), Dec 23 2001

Terms a(1) - a(20) are from the cited reference. The rest, a(21) - a(38) are obtained using the program from the same reference. Milan Stefanovic (stefke381(AT)gmail.com), Nov 19 2008

Broken link corrected by Rainer Rosenthal (r.rosenthal(AT)web.de), Jun 18 2009