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 A271216 a(n) = 2^n floor(n/2)! 2

%I #5 Nov 07 2016 09:05:31

%S 1,2,4,8,32,64,384,768,6144,12288,122880,245760,2949120,5898240,

%T 82575360,165150720,2642411520,5284823040,95126814720,190253629440,

%U 3805072588800

%N a(n) = 2^n floor(n/2)!

%C Number of symmetric rearrangement maps, i.e., rearrangement maps which satisfy a=a^(AI) and a^A = a^I.

%D J. Burns, Counting a Class of Signed Permutations and Chord Diagrams related to DNA Rearrangement, Preprint.

%H J. Burns, <a href="http://jtburns.myweb.usf.edu/tables/rearrangement_maps.html">Table of Rearrangement Maps and Patterns for n = 1, 2, and 3</a>.

%F a(n) = 2^n floor(n/2)!

%e For n=0 the a(0)=1 solution is { ∅ }

%e For n=1 the a(1)=2 solutions are { +1, -1 }

%e For n=2 the a(2)=4 solutions are { +1+2, -2-1, +2+1, -1-2 }

%e For n=3 the a(3)=8 solutions are { +1+2+3, -3-2-1, +3-2+1, -1+2-3, +3+2+1, -1-2-3, +1-2+3, -3+2-1 }

%t Table[2^n*Floor[n/2]!,{n,0,20}]

%Y Cf. A000165, A271213.

%K nonn,easy

%O 0,2

%A _Jonathan Burns_, Apr 13 2016

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