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Search: id:A078601
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| A078601 |
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Number of ways to lace a shoe that has n pairs of eyelets, assuming the lacing satisfies certain conditions. |
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+0
3
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| 1, 3, 42, 1080, 51840, 3758400, 382838400, 52733721600, 9400624128000, 2105593491456000, 579255485276160000, 191957359005941760000, 75420399121328701440000, 34668462695110852608000000, 18432051070888873171353600000, 11223248177765618214764544000000, 7759395812038133743242706944000000
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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The lace must follow a Hamiltonian path through the 2n eyelets. At least one of the neighbors of every eyelet must be on the other side of the shoe.
The lace is "undirected": reversing the order of eyelets along the path does not count as a different solution.
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REFERENCES
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B. Polster, What is the best way to lace your shoes?, Nature, 420 (Dec 05 2002), 476.
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LINKS
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Index entries for sequences related to shoe lacings
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FORMULA
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a(1)=1; for n > 1, a(n) = ((n!)^2/2)*Sum(binomial(n-k, k)^2/(n-k), k=0..floor(n/2)).
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EXAMPLE
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Label the eyelets 1, ..., n from front to back on the left and from n+1, ..., 2n from back to front on the right. For n=2 the three solutions are 1 2 3 4, 3 1 2 4, 1 3 2 4.
For n=3 the first few solutions are 2 4 1 3 5 6, 1 4 2 3 5 6, 2 1 4 3 5 6, 1 2 4 3 5 6, 1 3 4 2 5 6, 3 1 4 2 5 6, 1 4 3 2 5 6, 3 4 1 2 5 6, 3 4 2 1 5 6, 2 4 3 1 5 6, 3 2 4 1 5 6, 2 3 4 1 5 6, 2 3 5 1 4 6, 3 2 5 1 4 6, 2 5 3 1 4 6, 3 5 2 1 4 6, ...
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MAPLE
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A078601 := n->((n!)^2/2)*add(binomial(n-k, k)^2/(n-k), k=0..floor(n/2));
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CROSSREFS
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See A078602 and A078629 for other ways of counting lacings.
Adjacent sequences: A078598 A078599 A078600 this_sequence A078602 A078603 A078604
Sequence in context: A003770 A097068 A092470 this_sequence A083402 A058808 A137192
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KEYWORD
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nonn
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AUTHOR
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njas, Dec 11 2002
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