%I #41 Feb 24 2019 01:35:47
%S 0,1,3,21,207,5559,299599,33562695,7635498335,3518440564543,
%T 3275345183542207,6148914696963883711,23248573454127484129023,
%U 176848577040808821410837119,2704321280486889389864215362559,83076749736557243209409446411255935,5124252113632955685095523500148980125695,634332307869315502692705867068871886072665599,157534492276509956656902449336997243381860518317055
%N 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 nonempty subset of these lines (subsets differing by a rotation are regarded as identical). a(1)=0, a(2)=1 by convention.
%C This is the number of ways to sew on a button having n holes arranged in a regular polygon. A button with no stitches would fall off, which is why we require that the subset be nonempty (cf. A192332).
%C A. Kolmogorov posed the problem for n=4 at the age of 6.
%D A. N. Shiryaev 'Andrei Nikolaevich Kolmogorov: A Biographical Sketch of His Life and Creative Paths' in Harold H. McFaden (translator), Kolmogorov in Perspective, American Mathematical Society (2000), page 4.
%H Vincenzo Librandi, <a href="/A192314/b192314.txt">Table of n, a(n) for n = 1..80</a>
%F a(p) = ((2^(p*(p-1)/2) - 2^((p-1)/2)) / p) + (2^((p-1)/2)) - 1 if p is prime.
%F Comment from _N. J. A. Sloane_, Jun 28 2011: More generally, 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))) - 1.
%p with(numtheory); f:=proc(n) local t0,t1,d; t0:=0; t1:=divisors(n);
%p for d in t1 do
%p if d mod 2 = 0 then t0:=t0+phi(d)*2^(n^2/(2*d))
%p else t0:=t0+phi(d)*2^(n*(n-1)/(2*d)); fi; od; t0/n; end;
%p [seq(f(n)-1, n=1..30)];
%p # _N. J. A. Sloane_, Jun 28 2011
%t Table[ -1 + 1/n * Plus @@ Map[Function[d, EulerPhi[d]*2^((n*(n - Mod[d, 2])/2)/d)], Divisors[n]], {n, 1, 20}] (* _Olivier GĂ©rard_, Aug 27 2011 *)
%Y Cf. A192332, A191563.
%K nonn
%O 1,3
%A _David N Lumsden_, Jun 27 2011
%E Terms from a(8) onwards from _N. J. A. Sloane_, Jun 28 2011