

A050998


Inequivalent solutions to Langford (or LangfordSkolem) 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.


5



231213, 23421314, 14156742352637, 14167345236275, 15146735423627, 15163745326427, 15167245236473, 15173465324726, 16135743625427, 16172452634753, 17125623475364, 17126425374635, 23627345161475
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OFFSET

1,1


COMMENTS

Entries are indexed by numbers n == 1 or 0 mod 4 (A014601).
More precisely, for each given n = (3, 4, 7, 8, ...) in A014601, all of the A014552(n) inequivalent solutions are listed in lexicographic order. For example, a(1), a(2) and a(3) correspond to n=3, 4 and 7, but a(4) is not the first solution for n=8 but the second solution for n=7.  M. F. Hasler, Nov 12 2015
"Inequivalent" means that for two solutions related by symmetry (reading the digits backwards), only the (lexicographic) smaller one is listed.  M. F. Hasler, Nov 15 2015
It is unclear how the sequence goes on after the first 1+1+26+150 terms, with the solutions for n >= 11. Will a solution s=(s[1],...,s[n]) be coded again by Sum_{i=1..n} s[i]*b^(ni) in base b=10, or in some larger base b >= n+1? Maybe using as many decimal digits as needed, i.e., b=100 for 11 <= n <= 99?  M. F. Hasler, Nov 16 2015


REFERENCES

M. Gardner, Mathematical Magic Show, New York: Vintage, pp. 70 and 7778, 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 129159, 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.


EXAMPLE

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).


CROSSREFS

See A014552 (the main entry for this problem) for number of solutions.
Sequence in context: A015319 A270515 A179624 * A237462 A244349 A128484
Adjacent sequences: A050995 A050996 A050997 * A050999 A051000 A051001


KEYWORD

nonn,nice,easy


AUTHOR

Eric W. Weisstein


EXTENSIONS

Definition clarified by M. F. Hasler, Nov 15 2015


STATUS

approved



