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%I #29 Aug 10 2019 12:47:00
%S 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
%N Number of solutions to triples version of Langford (or Langford-Skolem) problem.
%C 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?
%C a(n)=0 for n mod 9 not in {-1,0,1}. - _Gheorghe Coserea_, Aug 23 2017
%H F. S. Gillespie and W. R. Utz, <a href="http://www.fq.math.ca/Scanned/4-2/gillespie.pdf">A generalized Langford Problem</a>, Fibonacci Quart., 1966, v4, 184-186.
%H J. E. Miller, <a href="http://dialectrix.com/langford.html">Langford's Problem</a>
%e For n=9 the a(9)=3 solutions, up to reversal of the order, are:
%e 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
%e 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
%e 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
%e From _Gheorghe Coserea_, Aug 26 2017: (Start)
%e For n=10 the a(10)=5 solutions, up to reversal of the order, are:
%e 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
%e 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
%e 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
%e 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
%e 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
%e (End)
%Y Cf. A014552, A050998, A059106, A059108.
%K nonn,nice,hard,more
%O 1,9
%A _N. J. A. Sloane_, Feb 14 2001