

A066385


Smallest maximum of sum of 3 consecutive terms in any arrangement of [1..n] in a circle.


0



6, 9, 10, 11, 14, 15, 16, 18, 20, 21, 23, 24, 25, 27, 29, 30, 32, 33, 35, 36, 38, 39, 41, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60
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OFFSET

3,1


COMMENTS

In a problem in the "Bundeswettbewerb 2002" competition there are 12 sticks of lengths 1,...,12 placed in a circle 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 ceiling(3*(n+1)/2), since n*a(n) >= Sum_{i=1..n} (p(i1)+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.
From Timothy Rolfe (trolfe(AT)ewu.edu), Jan 24 2010: (Start)
Here is the chronology of the clockface problem:
1. Dean S. Clark, "A Combinatorial Theorem on Circulant Matrices," American Mathematics Monthly, December 1985.
2. Martin Gardner's August 1986 column for Isaac Asimov's Science Fiction Magazine, which posed the problem of finding all legal permutations.
3. Timothy Rolfe, "Recurse Around the Clock", Mathematics and Computer Education, Vol. 21, No. 2 (Spring, 1987), pp. 98104.
4. In the Pacific Northwest regional contest as part of the ACM International Collegiate Programming Contest, the problem was problem E. Navigate from http://www.acmicpcpacnw.org/ProblemSet/2002/forweb.zip  specifically, ACMContest/2002/Contest Problems/Final Versions/FinalContestFiles/e_CircPerm/
5. Timothy Rolfe, "Backtracking Algorithms", Dr. Dobb's Journal, Vol. 29, No. 5 (May 2004), pp. 48, 5051.
It was [5] that caused Paul W. Purdom, Jr., of Indiana University, Bloomington, to correspond with me, proposing a bounding function. This is what that generated our joint article. (End)
From Jon E. Schoenfield, Jun 24 2019: (Start)
Observations:
1. If we define f(n) = ceiling((3/2)*n) + 3, then a(n) = f(n) for all n < 39 except 3, 5, 6, 9, and 15.
2. a(n) = f(n) for all even n > 6.
3. f(n)1 <= a(n) <= f(n) for all odd n > 3.
Conjecture: for all odd n > 15, a(n) = f(n). (End)
From Jon E. Schoenfield, Jul 07 2019: (Start)
If f(n) is defined as above, then for all n >= 3, the following steps generate a permutation p of 1..n such that the maximum of the consecutive triple sums does not exceed f(n):
If n mod 9 = 0 then let t = n/3 and u = n/9; for each k in 1..n,
p(k) = ceiling(n  (t + 1/2)*r  (3/2)*s) if s < 2*u,
1 + 3*s  (t1)*r otherwise,
where r = k mod 3
and s = (floor(k/3)  2*r*u) mod t.
If n mod 9 = 3 or 6 then let t = n/3; for each k in 1..n,
p(k) = ceiling(n  t*r  (3/2)*s) if s < 2*t/3,
1 + 3*s  t*r otherwise,
where r = k mod 3
and s = (floor(k/3)  ((t*(t mod 3)  1)*r)/3) mod t.
If n mod 3 != 0 then for each k in 1..n,
p(k) = k+1 if k == n+1 (mod 3),
ceiling((n*(((k mod 3) mod 2) + 1)  k) / 2) otherwise.
(The maximum of the resulting consecutive triple sums is equal to f(n) for all n >= 3 except at n = 3, 6, 9, and 15; at those values of n, that maximum is less than f.)
It is not difficult to prove that, for even values of n > 6, a(n) >= (3/2)*n + 3; that lower bound, together with the upper bound provided by the steps above, demonstrates that a(n) = (3/2)*n + 3 for all even n > 6.
It seems to me that there should be a way to prove that, for all odd values of n > 15, a(n) must exceed (3/2)*n + 5/2. If such a proof were found, then it, together with the upper bound provided by the steps above, would demonstrate that a(n) = (3/2)*n + 7/2 for all odd n > 15, and thus a(n) = f(n) = ceiling((3/2)*n) + 3 for all n > 15. (End)


REFERENCES

de.sci.mathematik, Thread "Zahlenkreis", December 2001
Martin Gardner's August 1986 column for Isaac Asimov's Science Fiction Magazine, which posed the problem of finding all legal permutations.
Timothy Rolfe, "Backtracking Algorithms", Dr. Dobb's Journal, Vol. 29, No. 5 (May 2004), pp. 48, 5051.


LINKS

Table of n, a(n) for n=3..38.
ACM International Collegiate Programming Contest, Pacific Northwest Regional Contest, Problem E, then specifically, ACMContest/2002/Contest Problems/Final Versions/FinalContestFiles/e_CircPerm/.
Bundeswettbewerb Mathematik, Problem 2002 (1.4) [Broken link?]
Dean S. Clark, A Combinatorial Theorem on Circulant Matrices, American Mathematics Monthly, December 1985.
Timothy Rolfe, Recurse Around the Clock, Mathematics and Computer Education, Vol. 21, No. 2 (Spring, 1987), pp. 98104.
Timothy Rolfe, Backtracking Algorithms, Dr. Dobb's website, May 01 2004.
T. J. Rolfe and 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. 8384 [From Milan Stefanovic (stefke381(AT)gmail.com), Nov 19 2008]


FORMULA

Let p be a permutation of 1..n and let g(p) be the maximum of the consecutive triple sums p(i1)+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.


EXAMPLE

a(6)=11 because the cycle 145236 has sums 11,10,11,10,11,10 with max=11, and there is no cycle that yields a smaller maximum.
This example by Helmut Richter shows that a(14) = 24 is very likely: p = (1811491021256133714) with g(p) = 11+4+9 = 24 as maximal threesum.


MATHEMATICA

(* *Warning* This is a heuristic program. Output data are not reliable beyond a(38) *) a[n_] := a[n] = If[n < 17, {6, 9, 10, 11, 14, 15, 16, 18, 20, 21, 23, 24, 25, 27}[[n2]], 3n + 5  a[n  1] ]; Table[a[n], {n, 3, 38}] (* JeanFrançois Alcover, Mar 13 2016, using "FindSequenceFunction" *)


CROSSREFS

Sequence in context: A070598 A262435 A124257 * A226913 A293826 A103092
Adjacent sequences: A066382 A066383 A066384 * A066386 A066387 A066388


KEYWORD

nice,nonn,more


AUTHOR

Rainer Rosenthal, Dec 23 2001


EXTENSIONS

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, Jun 18 2009


STATUS

approved



