

A059106


Number of solutions to Nickerson variant of Langford (or LangfordSkolem) problem.


9



1, 0, 0, 3, 5, 0, 0, 252, 1328, 0, 0, 227968, 1520280, 0, 0, 700078384, 6124491248, 0, 0, 5717789399488, 61782464083584, 0, 0, 102388058845620672, 1317281759888482688, 0, 0
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OFFSET

1,4


COMMENTS

How many ways are of arranging the numbers 1,1,2,2,3,3,...,n,n so that there are zero numbers between the two 1's, one number between the two 2's, ..., n1 numbers between the two n's?
For n>1, a(n) = A004075(n)/2 because A004075 also counts reflected solutions.  Martin Fuller, Mar 08 2007
Because of symmetry, is a(5) = 5 the largest prime in this sequence?  Jonathan Vos Post, Apr 02 2011


LINKS

Table of n, a(n) for n=1..27.
Ali Assarpour, Amotz BarNoy, Ou Liuo, Counting the Number of Langford Skolem Pairings, arXiv:1507.00315 [cs.DM], 2015.
Gheorghe Coserea, Solutions for n=8.
Gheorghe Coserea, Solutions for n=9.
J. E. Miller, Langford's Problem
R. S. Nickerson and D. C. B. Marsh, E1845: A variant of Langford's Problem, American Math. Monthly, 1967, 74, 591595.


EXAMPLE

For n=4 the a(4)=3 solutions, up to reversal of the order, are:
1 1 3 4 2 3 2 4
1 1 4 2 3 2 4 3
2 3 2 4 3 1 1 4
From Gheorghe Coserea, Aug 26 2017: (Start)
For n=5 the a(5)=5 solutions, up to reversal of the order, are:
1 1 3 4 5 3 2 4 2 5
1 1 5 2 4 2 3 5 4 3
2 3 2 5 3 4 1 1 5 4
2 4 2 3 5 4 3 1 1 5
3 5 2 3 2 4 5 1 1 4
(End)


CROSSREFS

Cf. A014552, A050998, A059107, A059108.
Cf. A004075, A268535.
Sequence in context: A230424 A113037 A063866 * A087676 A291207 A058813
Adjacent sequences: A059103 A059104 A059105 * A059107 A059108 A059109


KEYWORD

nonn,nice,hard,more


AUTHOR

N. J. A. Sloane, Feb 14 2001


EXTENSIONS

a(20)a(23) from Mike Godfrey (m.godfrey(AT)umist.ac.uk), Mar 14 2002
Extended using results from the Assarpour et al. (2015) paper by N. J. A. Sloane, Feb 22 2016 at the suggestion of William Rex Marshall


STATUS

approved



