|
%I #7 Aug 21 2017 17:24:25
%S 1,1,1,4,9,64,25,256,1225,16384,3969,65536,53361,1048576,184041,
%T 4194304,41409225,1073741824,147744025,4294967296,2133423721,
%U 68719476736,7775536041,274877906944,457028729521,17592186044416,1690195005625
%N Numerators of ratio of sides of n-th triple of rectangles of unit area sum around a triangle.
%C Page 13 of the link shows the type of configuration. When n is odd, the numerators 1,1,9,25,1225,3969,.. are A038534 and (A001790)^2, and the denominators 1,4,64,256,16384,65536,.. are A056982, A038533/2, and (A046161)^2. When n is even, the numerators 1,4,64,256,16384,65536,.. are A056982, A038533/2, and (A046161)^2, and the denominators 3,27,675,3675,297675,1440747,.. are 3*(A001803)^2. The limit of a(n+1)/a(n) as n(odd) tends to infinity = Pi^2/12, A072691. The limit of a(n+2)/a(n) as n tends to infinity = 1. a(n), for large odd n, tends to 2/(Pi*n). a(n), for large even n, tends to Pi/(6*n). The expansion of 2*x*EllipticK(x)/Pi gives the odd fractions. The expansion of 1/3*x*HypergeometricPFQ({1,1,1},{3/2,3/2},x) gives the even fractions.
%H Paul Yiu, <a href="http://math.fau.edu/yiu/EuclideanGeometryNotes.pdf">EuclideanGeometry Notes</a>
%F a(n)=a(n-2)*((n-2)/(n-1))^2, a(1)=1, a(2)=1/3. a(n)=((n/2-1)!)^2/(Pi*((n/2-1/2)!)^2) for n odd. a(n)=(2^(1-n)*(n-2)!!^2)/((n-1)/2)!^2 for n odd. a(n)=Pi*((n/2-1)!)^2/(12*((n/2-1/2)!)^2) for n even. a(n)=(2^(n-2)*((n-2)/2)!^2)/(3*(n-1)!!^2) for n even.
%e a(5) = a(5-2)*((5-2)/(5-1))^2 = 1/4*(3/4)^2 = 9/64
%t a[n_]:=If[OddQ[n], ((n/2-1)!)^2/(Pi*((n/2-1/2)!)^2), Pi*((n/2-1)!)^2/(12*((n/2-1/2)!)^2)] a[n_]:=If[OddQ[n], (2^(1-n)*(n-2)!!^2)/((n-1)/2)!^2, (2^(n-2)*((n-2)/2)!^2)/(3*(n-1)!!^2)] a[n_]:=((12+Pi^2+E^(I*n*Pi)*(Pi^2-12))*((n/2-1)!)^2)/(24*Pi*((n/2-1/2)!)^2) (CoefficientList[Series[(I*x*(6+Sqrt[3]*Pi)-2*x*Sqrt[3]*Log[x+Sqrt[x^2-1]])/(6*Sqrt[x^2-1]), {x, 0, 20}], x])^2
%Y Cf. A005386, A092936, A094084.
%K easy,frac,nonn
%O 1,4
%A _Peter J. C. Moses_, Apr 30 2004
|
