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A059107
Number of solutions to triples version of Langford (or Langford-Skolem) problem.
3
0, 0, 0, 0, 0, 0, 0, 0, 3, 5, 0, 0, 0, 0, 0, 0, 13440, 54947, 249280, 0, 0, 0, 0, 0, 0
OFFSET
1,9
COMMENTS
How many ways are there of arranging the numbers 1,1,1,2,2,2,3,3,3, ...,n,n,n so that there is one number between the first and second 1's and one number between the second and third 1's; two numbers between the first and second 2's and two numbers between the second and third 2's; ... n numbers between the first and second n's and n numbers between the second and third n's?
a(n)=0 for n mod 9 not in {-1,0,1}. - Gheorghe Coserea, Aug 23 2017
LINKS
F. S. Gillespie and W. R. Utz, A generalized Langford Problem, Fibonacci Quart., 1966, v4, 184-186.
J. E. Miller, Langford's Problem
EXAMPLE
For n=9 the a(9)=3 solutions, up to reversal of the order, are:
1 8 1 9 1 5 2 6 7 2 8 5 2 9 6 4 7 5 3 8 4 6 3 9 7 4 3
1 9 1 2 1 8 2 4 6 2 7 9 4 5 8 6 3 4 7 5 3 9 6 8 3 5 7
1 9 1 6 1 8 2 5 7 2 6 9 2 5 8 4 7 6 3 5 4 9 3 8 7 4 3
From Gheorghe Coserea, Aug 26 2017: (Start)
For n=10 the a(10)=5 solutions, up to reversal of the order, are:
1 3 1 10 1 3 4 9 6 3 8 4 5 7 10 6 4 9 5 8 2 7 6 2 5 10 2 9 8 7
1 10 1 2 1 4 2 9 7 2 4 8 10 5 6 4 7 9 3 5 8 6 3 10 7 5 3 9 6 8
1 10 1 6 1 7 9 3 5 8 6 3 10 7 5 3 9 6 8 4 5 7 2 10 4 2 9 8 2 4
4 10 1 7 1 4 1 8 9 3 4 7 10 3 5 6 8 3 9 7 5 2 6 10 2 8 5 2 9 6
5 2 7 9 2 10 5 2 6 4 7 8 5 9 4 6 10 3 7 4 8 3 6 9 1 3 1 10 1 8
(End)
CROSSREFS
KEYWORD
nonn,nice,hard,more
AUTHOR
N. J. A. Sloane, Feb 14 2001
STATUS
approved