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

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

Table of n, a(n) for n=1..25.

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

Cf. A014552, A050998, A059106, A059108.

Sequence in context: A094396 A161838 A152624 * A025115 A230424 A113037

Adjacent sequences: A059104 A059105 A059106 * A059108 A059109 A059110

Keyword

nonn,nice,hard,more

Author

N. J. A. Sloane, Feb 14 2001

Status

approved