A078601 Number of ways to lace a shoe that has n pairs of eyelets, assuming the lacing satisfies certain conditions.

1, 3, 42, 1080, 51840, 3758400, 382838400, 52733721600, 9400624128000, 2105593491456000, 579255485276160000, 191957359005941760000, 75420399121328701440000, 34668462695110852608000000, 18432051070888873171353600000, 11223248177765618214764544000000, 7759395812038133743242706944000000

Offset: 1

Author

N. J. A. Sloane, Dec 11 2002

Keywords

• nonn

Comments

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.

Examples

			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, ...
		

Links

• B. Polster, What is the best way to lace your shoes?, Nature, 420 (Dec 05 2002), 476.
• Index entries for sequences related to shoe lacings

Crossrefs

See A078602 and A078629 for other ways of counting lacings.

Cf. A123385.

Programs

Maple

A078601 := n->((n!)^2/2)*add(binomial(n-k,k)^2/(n-k),k=0..floor(n/2));

Mathematica

a[n_] := If[n == 1, 1, n!^2/2 Sum[Binomial[n-k, k]^2/(n-k), {k, 0, n/2}]];
a /@ Range[1, 17] (* Jean-François Alcover, Oct 01 2019 *)

PARI

a(n)=if(n>1,n!^2*sum(k=0,n\2,binomial(n-k, k)^2/(n-k))/2,1) \\ Charles R Greathouse IV, Sep 10 2015

Python

from sympy import factorial, binomial
a = lambda n:((factorial(n)**2)>>1) * sum((binomial(n-k,k)**2)/(n-k) for k in range(0,(n>>1)+1)) if n > 1 else 1
print([a(n) for n in range(1, 18)]) # Darío Clavijo, Mar 06 2024

Formula

a(1)=1; for n > 1, a(n) = ((n!)^2/2)*Sum_{k=0..floor(n/2)} binomial(n-k, k)^2/(n-k).