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Search: id:A059107
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| A059107 |
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Number of solutions to triples version of Langford (or Langford-Skolem) problem. |
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4
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| 0, 0, 0, 0, 0, 0, 0, 0, 3, 5, 0, 0, 0, 0, 0, 0, 13440, 54947, 249280, 0
(list; graph; listen)
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OFFSET
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1,9
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COMMENT
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How many ways are 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?
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REFERENCES
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Gillespie and Utz, A generalized Langford Problem, Fibonacci Quart., 1966, v4, 184-186.
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LINKS
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J. E. Miller, Langford's Problem
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EXAMPLE
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For n=9 a solution is 3 4 7 9 3 6 4 8 3 5 7 4 6 9 2 5 8 2 7 6 2 5 1 9 1 8 1.
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CROSSREFS
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Cf. A014552, A050998, A059106, A059108.
Sequence in context: A042711 A069970 A094396 this_sequence A025115 A113037 A063866
Adjacent sequences: A059104 A059105 A059106 this_sequence A059108 A059109 A059110
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KEYWORD
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nonn,nice,hard
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AUTHOR
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njas, Feb 14 2001
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