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A059106 Number of solutions to Nickerson variant of Langford (or Langford-Skolem) 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, 3532373626038214732032, 52717585747603598276736, 0, 0 (list; graph; refs; listen; history; text; internal format)
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, ..., n-1 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
Ali Assarpour, Amotz Bar-Noy, 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, 591-595.
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
Sequence in context: A230424 A113037 A063866 * A318521 A087676 A291207
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
a(28)-a(31) from Assarpour et al. (2015), added by Max Alekseyev, Sep 24 2023
STATUS
approved

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Last modified April 25 05:18 EDT 2024. Contains 371964 sequences. (Running on oeis4.)