cp's OEIS Frontend

A059107 Number of solutions to triples version of Langford (or Langford-Skolem) problem.

%I A059107 #31 Jan 05 2025 19:51:36
%S A059107 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 A059107 Number of solutions to triples version of Langford (or Langford-Skolem) problem.
%C A059107 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 A059107 a(n)=0 for n mod 9 not in {-1,0,1}. - _Gheorghe Coserea_, Aug 23 2017
%H A059107 F. S. Gillespie and W. R. Utz, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/4-2/gillespie.pdf">A generalized Langford Problem</a>, Fibonacci Quart., 1966, v4, 184-186.
%H A059107 J. E. Miller, <a href="http://dialectrix.com/langford.html">Langford's Problem</a>
%e A059107 For n=9 the a(9)=3 solutions, up to reversal of the order, are:
%e A059107 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 A059107 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 A059107 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 A059107 From _Gheorghe Coserea_, Aug 26 2017: (Start)
%e A059107 For n=10 the a(10)=5 solutions, up to reversal of the order, are:
%e A059107 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 A059107 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 A059107 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 A059107 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 A059107 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 A059107 (End)
%Y A059107 Cf. A014552, A050998, A059106, A059108.
%K A059107 nonn,nice,hard,more
%O A059107 1,9
%A A059107 _N. J. A. Sloane_, Feb 14 2001