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%I #35 Sep 08 2022 08:45:26
%S 1,2,11,171,5736,312240,24327000,2554072920,347351195520,
%T 59397023589120,12473374574505600,3155763762320400000,
%U 946729128624509260800,332301924146113021900800,134914581203304233287756800,62735280259536165098353536000,33124227977035089658775531520000
%N Number of different convex inscribed polygons with n pair of sides of lengths d1, d2, ..., dn all distinct. Or number of bracelets with n pairs of beads, each pair of one among n colors.
%H G. C. Greubel, <a href="/A120445/b120445.txt">Table of n, a(n) for n = 1..235</a>
%H Ignacio Larrosa Cañestro, Marko Riedel, <a href="http://groups.google.com/group/es.ciencia.matematicas/msg/212060ce9e321055">n-digonos</a>.
%H Marko Riedel, <a href="http://math.stackexchange.com/questions/2024406/">Pairs of beads on a ring.</a>
%F a(n) = ((2n)!/2^n + (2n+1)*n!)/(4n).
%F a(n) ~ sqrt(Pi)*2^n*n^(2*n-1/2)/(2*exp(2*n)). - _Ilya Gutkovskiy_, Nov 21 2016
%e a(2) = 2 because there are two quadrilaterals with sides {1, 1, 2, 2}: a kite and a rectangle.
%t Table[((2 n)! / 2^n + (n + 1) n! + n n!) / (4 n), {n, 20}] (* _Vincenzo Librandi_, Nov 22 2016 *)
%o (Magma) [(Factorial(2*n)/2^n+(n+1)*Factorial(n)+n*Factorial(n))/(4*n): n in [1..20]]; // _Vincenzo Librandi_, Nov 22 2016
%o (PARI) for(n=1,25, print1(((2*n)!/2^n + (2*n+1)*n!)/(4*n), ", ")) \\ _G. C. Greubel_, May 25 2017
%K nonn
%O 1,2
%A _Ignacio Larrosa Cañestro_, Jul 19 2006
%E Name clarified by _Marko Riedel_, Nov 22 2016
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