A186783 Diameter of the symmetric group S_n when generated by the transposition (1,2) and both left and right rotations by (1,2,...,n).

0, 1, 2, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78

Offset: 1

Author

Tony Bartoletti, Feb 26 2011

Keywords

• nonn
• more
• hard

Comments

Given an ordered sequence of n elements (1,2,3,...,n), let X represent the permutation that transposes the first two elements, X(1,2,3,...,n) = (2,1,3,...,n), let L be the "left rotation" of the sequence, L(1,2,3,...,n) = (2,3,...,n,1), and let R be the "right rotation", R(1,2,3,...,n) = (n,1,2,...,n-1). Then every permutation of (1,2,3,...,n) can be expressed as a composition of the permutations X, L and R. One can exhaustively generate such compositions by taking L="0", X="1", R="2", and considering, in turn, base 3 numbers of increasing length (padded with leading zeros). Note that any base 3 number containing the subsequence "11", "02" or "20" may be discarded.

Note also that by defining the distance between any two permutations p and q in S_n, dist(p,q), to be the length of the minimal composition of LXR transforming p into q, we have dist(p,q) = dist(q,p), owing to L and R being mutually inverse, and X being self-inverse.

Conjecture from Chervov et. al., section 4.2: The longest and unique element of S_n is (1,2)(n,3)(n-1,4)(n-2,5)... = (2,1,n,n-1,n-2,...,3). The conjecture holds for n <= 13. - Dmitry Kamenetsky, Jun 22 2025

Examples

			The diameter of S_5 is 10, given this set of generators, since there is no sequence shorter than 0010010121 (i.e., LLXLLXLXRX) that will transform (1,2,3,4,5) into (2,1,5,4,3), and there is no permutation of (1,2,3,4,5) that requires more than a length-10 composition of L, X and R. Thus a(5) = 10.
		

Links

• Danilo Bazzanella, Antonio Di Scala, Simone Dutto, and Nadir Murru, Primality tests, linear recurrent sequences and the Pell equation, arXiv:2002.08062 [math.NT], 2020.
• A. Chervov, A. Soibelman, S. Lytkin, et al., CayleyPy RL: Pathfinding and Reinforcement Learning on Cayley Graphs, arXiv:2502.18663 [cs.LG], 2025. See pp. 1, 2, 4, 18, 24.
• Sai Satwik Kuppili and Bhadrachalam Chitturi, An Upper Bound for Sorting R_n with LRE, arXiv:2002.07342 [cs.DS], 2020.

Crossrefs

Cf. A039745, A186752, A000217.

Programs

Mathematica

a[1] = 0; a[n_] := GraphDiameter[CayleyGraph[PermutationGroup[{Cycles[{{1, 2}}], Cycles[{Range[n]}], InversePermutation[Cycles[{Range[n]}]]}]]]; (* Ben Whitmore, Jan 09 2018 *)

Sage

def a(n): return PermutationGroup([[(1, 2)], [tuple(1..n)], PermutationGroupElement([tuple(1..n)])^(-1)]).cayley_graph().diameter() # Max Alekseyev, Sep 09 2011

Formula

Conjecture: a(n) = - Sum_{k=1..n-1} Stirling1(n+k-1, (n-1)*k). This formula holds for all known n. - Arkadiusz Wesolowski, Mar 30 2013. For n>3, this formula contains only one nonzero term (for k=1) and reduces to the formula n*(n-1)/2 conjectured below. - Max Alekseyev, Sep 10 2020

Conjecture: a(n) = n*(n-1)/2 for all n>3. - Per W. Alexandersson, Aug 25 2020

Conjectures from Colin Barker, Aug 26 2020: (Start)

G.f.: x^2*(1 - x + 3*x^2 - 3*x^3 + x^4) / (1 - x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

(End)

Extensions

a(8)=28 added by Tony Bartoletti, Mar 12 2011

a(9)=36 added by R. H. Hardin, Sep 09 2011

a(10)=45 added by Sharon Li, Mar 09 2013

a(11)=55 and a(12)=66 added by James Bieron, Mar 15 2013

a(13)=78 added by Ben Whitmore, Jan 09 2018