A050998 Inequivalent solutions to Langford (or Langford-Skolem) problem of arranging the numbers 1,1,2,2,3,3,...,n,n so that there is one number between the two 1's, two numbers between the two 2's, ..., n numbers between the two n's, listed by length and lexicographic order.
231213, 23421314, 14156742352637, 14167345236275, 15146735423627, 15163745326427, 15167245236473, 15173465324726, 16135743625427, 16172452634753, 17125623475364, 17126425374635, 23627345161475
Offset: 1
Examples
The first n which allows a solution (A014552(n) > 0; n in A014601) is n=3, the solutions are a(1) = 231213 and the same read backwards, 312132. The next solutions are given for n=4, again there is only A014552(4)=1 solution a(2) = 23421314 up to reversal (41312432, not listed). Then follow the A014552(7)=26 (inequivalent) solutions for n=7, viz. a(3)-a(28).
References
- M. Gardner, Mathematical Magic Show, New York: Vintage, pp. 70 and 77-78, 1978.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..178
- R. K. Guy, The unity of combinatorics, Proc. 25th Iranian Math. Conf, Tehran, (1994), Math. Appl 329 129-159, Kluwer Dordrecht 1995, Math. Rev. 96k:05001.
- C. D. Langford, Problem, Math. Gaz., 1958, vol. 42, p. 228.
- J. Miller, Langford's problem.
- Eric Weisstein's World of Mathematics, Langford's Problem.
Crossrefs
See A014552 (the main entry for this problem) for number of solutions.
Extensions
Definition clarified by M. F. Hasler, Nov 15 2015
Comments