login
Number of solutions to variant of triples version of Langford (or Langford-Skolem) problem.
8

%I #37 Nov 22 2018 12:08:04

%S 1,1,0,0,0,0,0,0,0,9,20,33,0,0,0,0,0,0,200343,869006,4247790,0,0,0,0,

%T 0,0

%N Number of solutions to variant of triples version of Langford (or Langford-Skolem) problem.

%C How many ways are of arranging the numbers 1,1,1,2,2,2,3,3,3,...,n,n,n so that there are zero numbers between the first and second 1's and zero numbers between the second and third 1's; one number between the first and second 2's and one number between the second and third 2's; ... n-1 numbers between the first and second n's and n-1 numbers between the second and third n's?

%C a(n)=0 for n mod 9 not in {0,1,2}. - _Gheorghe Coserea_, Aug 23 2017

%H Gheorghe Coserea, <a href="/A059108/a059108.txt">Solutions for n=10</a>.

%H Gheorghe Coserea, <a href="/A059108/a059108_1.txt">Solutions for n=11</a>.

%H J. E. Miller, <a href="http://dialectrix.com/langford.html">Langford's Problem</a>

%e From _Gheorghe Coserea_, Jul 14 2017: (Start)

%e For n=9 the a(9)=9 solutions, up to reversal of the order, are:

%e 2 4 2 8 2 4 6 7 9 4 3 8 6 3 7 5 3 9 6 8 5 7 1 1 1 5 9

%e 2 4 2 9 2 4 5 6 7 4 8 5 9 6 3 7 5 3 8 6 3 9 7 1 1 1 8

%e 4 2 5 2 4 2 9 5 4 7 8 3 5 6 3 9 7 3 8 6 1 1 1 7 9 6 8

%e 5 1 1 1 7 5 8 6 9 3 5 7 3 6 8 3 4 9 7 6 4 2 8 2 4 2 9

%e 5 6 1 1 1 5 8 6 9 3 5 7 3 6 8 3 4 9 7 2 4 2 8 2 4 7 9

%e 6 7 9 2 5 2 6 2 7 5 8 9 6 3 5 7 3 4 8 3 9 4 1 1 1 4 8

%e 6 7 9 2 5 2 6 2 7 5 8 9 6 4 5 7 3 4 8 3 9 4 3 1 1 1 8

%e 7 4 2 8 2 4 2 7 9 4 3 8 6 3 7 5 3 9 6 8 5 1 1 1 6 5 9

%e 7 5 3 6 9 3 5 7 3 6 8 5 4 9 7 6 4 2 8 2 4 2 9 1 1 1 8

%e (End)

%Y Cf. A014552, A050998, A059106, A059107.

%K nonn,nice,hard,more

%O 0,10

%A _N. J. A. Sloane_, Feb 14 2001

%E _Fausto A. C. Cariboni_ has confirmed the values a(1) to a(20). - _N. J. A. Sloane_, Mar 27 2017

%E a(21) from _Fausto A. C. Cariboni_, Mar 28 2017