A192332 For n >= 3, draw a regular n-sided polygon and its n(n-3)/2 diagonals, so there are n(n-1)/2 lines; a(n) is the number of ways to choose a subset of these lines (subsets differing by a rotation are regarded as identical). a(1)=1, a(2)=2 by convention.
1, 2, 4, 22, 208, 5560, 299600, 33562696, 7635498336, 3518440564544, 3275345183542208, 6148914696963883712, 23248573454127484129024, 176848577040808821410837120, 2704321280486889389864215362560, 83076749736557243209409446411255936, 5124252113632955685095523500148980125696, 634332307869315502692705867068871886072665600
Offset: 1
Keywords
Examples
From _Gus Wiseman_, Mar 04 2019: (Start) Inequivalent representatives of the a(1) = 1 through a(4) = 22 graphical necklace edge-sets: {} {} {} {} {{12}} {{12}} {{12}} {{12}{13}} {{13}} {{12}{13}{23}} {{12}{13}} {{12}{14}} {{12}{24}} {{12}{34}} {{13}{24}} {{12}{13}{14}} {{12}{13}{23}} {{12}{13}{24}} {{12}{13}{34}} {{12}{14}{23}} {{12}{24}{34}} {{12}{13}{14}{23}} {{12}{13}{14}{24}} {{12}{13}{14}{34}} {{12}{13}{24}{34}} {{12}{14}{23}{34}} {{12}{13}{14}{23}{24}} {{12}{13}{14}{23}{34}} {{12}{13}{14}{23}{24}{34}} (End)
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..80
- Gus Wiseman, Inequivalent representatives of the a(5) = 22 graphical necklaces.
- Gus Wiseman, Inequivalent representatives of the a(5) = 208 graphical necklaces.
Crossrefs
Programs
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Maple
with(numtheory); f:=proc(n) local t0, t1, d; t0:=0; t1:=divisors(n); for d in t1 do if d mod 2 = 0 then t0:=t0+phi(d)*2^(n^2/(2*d)) else t0:=t0+phi(d)*2^(n*(n-1)/(2*d)); fi; od; t0/n; end; [seq(f(n), n=1..30)];
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Mathematica
Table[ 1/n* Plus @@ Map[Function[d, EulerPhi[d]*2^((n*(n - Mod[d, 2])/2)/d)], Divisors[n]], {n, 1, 20}] (* Olivier Gérard, Aug 27 2011 *) rotgra[g_,m_]:=Sort[Sort/@(g/.k_Integer:>If[k==m,1,k+1])]; Table[Length[Select[Subsets[Subsets[Range[n],{2}]],#=={}||#==First[Sort[Table[Nest[rotgra[#,n]&,#,j],{j,n}]]]&]],{n,0,5}] (* Gus Wiseman, Mar 04 2019 *)
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PARI
a(n) = sumdiv(n, d, if (d%2, eulerphi(d)*2^(n*(n-1)/(2*d)), eulerphi(d)*2^(n^2/(2*d))))/n; \\ Michel Marcus, Mar 08 2019
Formula
a(n) = (1/n)*(Sum_{d|n, d odd} phi(d)*2^(n*(n-1)/(2*d)) + Sum_{d|n, d even} phi(d)*2^(n^2/(2*d))).
Comments