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A078700 Number of symmetric ways to lace a shoe that has n pairs of eyelets such that each eyelet has at least one direct connection to the opposite side. 6

%I #21 Sep 10 2017 05:19:02

%S 1,2,6,30,192,1560,15120,171360,2217600,32296320,522547200,9300614400,

%T 180583603200,3798482688000,86044973414400,2088355965696000,

%U 54064489070592000,1487129136869376000,43312058119249920000

%N Number of symmetric ways to lace a shoe that has n pairs of eyelets such that each eyelet has at least one direct connection to the opposite side.

%C The lace must pass through each eyelet exactly one and must begin and end at the extreme pair of eyelets.

%H <a href="/index/La#lacings">Index entries for sequences related to shoe lacings</a>

%F a(n) = (n-1)!*Fibonacci(n+1) = A000142(n-1)*A000045(n+1). - Conjectured by _Vladeta Jovovic_, Jan 23 2005, proved by Antti Karttunen, Jan 06 2007.

%F Proof of Jovovic's conjecture (_Antti Karttunen_, Jan 06 2007): (Start)

%F Because of the symmetry and the beginning and ending conditions, we need to consider only n-1 eyelets on the other side. If considering only the distance from the starting and ending eyelets (the "level" of each eyelet-pair) through which the lace is traveling (but ignoring on which side it is), the lace will induce some permutation of {1..(n-1)} after it has visited half of the eyelets (and the remaining half of its route is wholly determined by the symmetry). This gives the factor (n-1)!. Independently of this, the condition that each eyelet has at least one direct connection to the opposite side, means that there is a simple bijection with binary strings of length n with no two consecutive 0's. I.e. we mark 0 when the lace stays on the same side and 1 when it crosses to the other side.

%F From the starting eyelet the lace can either cross to the other side (but not to the top one) or stay on the same side. However, after visiting half of the eylets, the lace MUST cross to the other side (on the same level), so this leaves n-1 eyelets where the choice is free, except that there can be no two consecutive 0's on the route. This gives the factor Fibonacci(n+1).

%F (End)

%e a(3) = 6: label the eyelets 1,2,3 from front to back on the left side then 4,5,6 from back to front on the right side. The symmetric lacings are: 124356 154326 153426 142536 145236 135246.

%p ZL:=[S, {a = Atom, b = Atom, S = Prod(X,Sequence(Prod(X,b))), X = Sequence(b,card <= 1)}, labelled]: seq(combstruct[count](ZL, size=n), n=0..18); # _Zerinvary Lajos_, Mar 26 2008

%t Table[Fibonacci[n + 2]*n!, {n, 0, 18}] (* _Zerinvary Lajos_, Jul 09 2009 *)

%o (Scheme:) (define (A078700 n) (* (A000142 (- n 1)) (A000045 (+ n 1))))

%Y Cf. A000045, A005922, A014417, A078676, A078698, A078702.

%K nonn

%O 1,2

%A _Hugo Pfoertner_, Dec 18 2002

%E Jovovic's conjecture proved and more terms as well as Scheme-code added by _Antti Karttunen_, Jan 02 2007

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