cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A014552 Number of solutions to Langford (or Langford-Skolem) problem (up to reversal of the order).

Original entry on oeis.org

0, 0, 1, 1, 0, 0, 26, 150, 0, 0, 17792, 108144, 0, 0, 39809640, 326721800, 0, 0, 256814891280, 2636337861200, 0, 0, 3799455942515488, 46845158056515936, 0, 0, 111683611098764903232, 1607383260609382393152, 0, 0
Offset: 1

Views

Author

John E. Miller (john@timehaven.us), Eric W. Weisstein, N. J. A. Sloane

Keywords

Comments

These are also called Langford pairings.
2*a(n) = A176127(n) gives the number of ways of arranging the numbers 1,1,2,2,...,n,n so that there is one number between the two 1's, two numbers between the two 2's, ..., n numbers between the two n's.
a(n) > 0 iff n == 0 or 3 (mod 4).

Examples

			Solutions for n=3 and 4: 312132 and 41312432.
Solution for n=16: 16, 14, 12, 10, 13, 5, 6, 4, 15, 11, 9, 5, 4, 6, 10, 12, 14, 16, 13, 8, 9, 11, 7, 1, 15, 1, 2, 3, 8, 2, 7, 3.
		

References

  • Jaromir Abrham, "Exponential lower bounds for the numbers of Skolem and extremal Langford sequences," Ars Combinatoria 22 (1986), 187-198.
  • M. Gardner, Mathematical Magic Show, New York: Vintage, pp. 70 and 77-78, 1978.
  • M. Gardner, Mathematical Magic Show, Revised edition published by Math. Assoc. Amer. in 1989. Contains a postscript on pp. 283-284 devoted to a discussion of early computations of the number of Langford sequences.
  • R. K. Guy, The unity of combinatorics, Proc. 25th Iranian Math. Conf, Tehran, (1994), Math. Appl 329 129-159, Kluwer Dordrecht 1995, Math. Rev. 96k:05001.
  • D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.1.1, p. 2.
  • M. Krajecki, Christophe Jaillet and Alain Bui, "Parallel tree search for combinatorial problems: A comparative study between OpenMP and MPI," Studia Informatica Universalis 4 (2005), 151-190.
  • Roselle, David P. Distributions of integers into s-tuples with given differences. Proceedings of the Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1971), pp. 31--42. Dept. Comput. Sci., Univ. Manitoba, Winnipeg, Man., 1971. MR0335429 (49 #211). - From N. J. A. Sloane, Jun 05 2012

Crossrefs

See A050998 for further examples of solutions.
If the zeros are omitted we get A192289.

Formula

a(n) = A176127(n)/2.

Extensions

a(20) from Ron van Bruchem and Mike Godfrey, Feb 18 2002
a(21)-a(23) sent by John E. Miller (john@timehaven.us) and Pab Ter (pabrlos(AT)yahoo.com), May 26 2004. These values were found by a team at Université de Reims Champagne-Ardenne, headed by Michael Krajecki, using over 50 processors for 4 days.
a(24)=46845158056515936 was computed circa Apr 15 2005 by the Krajecki team. - Don Knuth, Feb 03 2007
Edited by Max Alekseyev, May 31 2011
a(27) from the J. E. Miller web page "Langford's problem"; thanks to Eric Desbiaux for reporting this. - N. J. A. Sloane, May 18 2015. However, it appears that the value was wrong. - N. J. A. Sloane, Feb 22 2016
Corrected and extended using results from the Assarpour et al. (2015) paper by N. J. A. Sloane, Feb 22 2016 at the suggestion of William Rex Marshall.

A059106 Number of solutions to Nickerson variant of Langford (or Langford-Skolem) problem.

Original entry on oeis.org

1, 0, 0, 3, 5, 0, 0, 252, 1328, 0, 0, 227968, 1520280, 0, 0, 700078384, 6124491248, 0, 0, 5717789399488, 61782464083584, 0, 0, 102388058845620672, 1317281759888482688, 0, 0, 3532373626038214732032, 52717585747603598276736, 0, 0
Offset: 1

Views

Author

N. J. A. Sloane, Feb 14 2001

Keywords

Comments

How many ways are of arranging the numbers 1,1,2,2,3,3,...,n,n so that there are zero numbers between the two 1's, one number between the two 2's, ..., n-1 numbers between the two n's?
For n > 1, a(n) = A004075(n)/2 because A004075 also counts reflected solutions. - Martin Fuller, Mar 08 2007
Because of symmetry, is a(5) = 5 the largest prime in this sequence? - Jonathan Vos Post, Apr 02 2011

Examples

			For n=4 the a(4)=3 solutions, up to reversal of the order, are:
1 1 3 4 2 3 2 4
1 1 4 2 3 2 4 3
2 3 2 4 3 1 1 4
From _Gheorghe Coserea_, Aug 26 2017: (Start)
For n=5 the a(5)=5 solutions, up to reversal of the order, are:
1 1 3 4 5 3 2 4 2 5
1 1 5 2 4 2 3 5 4 3
2 3 2 5 3 4 1 1 5 4
2 4 2 3 5 4 3 1 1 5
3 5 2 3 2 4 5 1 1 4
(End)
		

Crossrefs

Extensions

a(20)-a(23) from Mike Godfrey (m.godfrey(AT)umist.ac.uk), Mar 14 2002
Extended using results from the Assarpour et al. (2015) paper by N. J. A. Sloane, Feb 22 2016 at the suggestion of William Rex Marshall
a(28)-a(31) from Assarpour et al. (2015), added by Max Alekseyev, Sep 24 2023

A059108 Number of solutions to variant of triples version of Langford (or Langford-Skolem) problem.

Original entry on oeis.org

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, 0, 0
Offset: 0

Views

Author

N. J. A. Sloane, Feb 14 2001

Keywords

Comments

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?
a(n)=0 for n mod 9 not in {0,1,2}. - Gheorghe Coserea, Aug 23 2017

Examples

			From _Gheorghe Coserea_, Jul 14 2017: (Start)
For n=9 the a(9)=9 solutions, up to reversal of the order, are:
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
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
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
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
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
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
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
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
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
(End)
		

Crossrefs

Extensions

Fausto A. C. Cariboni has confirmed the values a(1) to a(20). - N. J. A. Sloane, Mar 27 2017
a(21) from Fausto A. C. Cariboni, Mar 28 2017
Showing 1-3 of 3 results.