3,1

In a problem in the "Bundeswettbewerb 2001" 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

Bundeswettbewerb Mathematik 2001

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