|
%I #24 Mar 23 2019 12:38:08
%S 4,32,512,4096,131072,1048576,16777216,134217728,8589934592,
%T 68719476736,1099511627776,8796093022208,281474976710656,
%U 2251799813685248,36028797018963968,288230376151711744,36893488147419103232,295147905179352825856,4722366482869645213696,37778931862957161709568
%N Denominator of the probability that 2n-dimensional Gaussian random triangle has an obtuse angle.
%C Presumably this is the same as A093581? - _Andrew S. Plewe_, Apr 18 2007
%H Robert Israel, <a href="/A102557/b102557.txt">Table of n, a(n) for n = 1..830</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GaussianTrianglePicking.html">Gaussian Triangle Picking</a>
%F From _Robert Israel_, Sep 29 2016: (Start)
%F a(n) is the denominator of p(n) = Sum_{k=n..2n-1} binomial(2n-1,k) 3^(2n-k)/4^(2n-1).
%F -(6n+3)p(n)+(14n+11)p(n+1)-(8n+8)p(n+2)=0 for n >= 1.
%F G.f. of p(n): 3x(1-1/sqrt(4-3x))/(2-2x). (End)
%e 3/4, 15/32, 159/512, 867/4096, 19239/131072, 107985/1048576, ...
%p p:= gfun:-rectoproc({(-6*n-3)*v(n)+(14*n+11)*v(n+1)+(-8*n-8)*v(n+2), v(0) = 0, v(1) = 3/4, v(2) = 15/32},v(n),remember):
%p seq(denom(p(n)),n=1..50); # _Robert Israel_, Sep 29 2016
%t a[n_] := (3^n/4^(2n-1)) Binomial[2n-1, n] Hypergeometric2F1[1, 1-n, 1+n, -1/3] // Denominator; Array[a, 20] (* _Jean-François Alcover_, Mar 22 2019 *)
%o (PARI) a(n) = denominator(sum(k=n, 2*n-1, binomial(2*n-1,k)*3^(2*n-k)/4^(2*n-1))); \\ _Michel Marcus_, Mar 23 2019
%Y Cf. A093581, A102556, A102558, A102559.
%K nonn,frac
%O 1,1
%A _Eric W. Weisstein_, Jan 14 2005
|
