A102557 Denominator of the probability that 2n-dimensional Gaussian random triangle has an obtuse angle.

4, 32, 512, 4096, 131072, 1048576, 16777216, 134217728, 8589934592, 68719476736, 1099511627776, 8796093022208, 281474976710656, 2251799813685248, 36028797018963968, 288230376151711744, 36893488147419103232, 295147905179352825856, 4722366482869645213696, 37778931862957161709568

Offset: 1

Author

Eric W. Weisstein, Jan 14 2005

Keywords

• nonn
• frac

Comments

Presumably this is the same as A093581? - Andrew S. Plewe, Apr 18 2007

a(n) equals A093581(n) for n <= 55000. - G. C. Greubel, Oct 20 2024

Examples

			3/4, 15/32, 159/512, 867/4096, 19239/131072, 107985/1048576, ...
		

Links

• Robert Israel, Table of n, a(n) for n = 1..830
• Eric Weisstein's World of Mathematics, Gaussian Triangle Picking

Crossrefs

Cf. A093581, A102556, A102558, A102559, A378071.

Programs

Magma

A102557:= func< n | Power(2, 4*n-2-(&+Intseq(2*(n-1), 2))) >;
[A102557(n): n in [1..30]]; // G. C. Greubel, Oct 20 2024

Maple

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):
seq(denom(p(n)),n=1..50); # Robert Israel, Sep 29 2016
# Alternative, assuming offset 0:
a := n -> numer((4^(2*n+1)*n!^2)/((2*n+3)*(2*n)!)): # Peter Luschny, Dec 05 2024

Mathematica

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 *)

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

SageMath

def A102557(n): return pow(2, 4*n-2 - sum((2*n-2).digits(2)))
[A102557(n) for n in range(1,31)] # G. C. Greubel, Oct 20 2024

Formula

From Robert Israel, Sep 29 2016: (Start)

a(n) is the denominator of p(n) = Sum_{k=n..2n-1} binomial(2n-1,k) 3^(2n-k)/4^(2n-1).

8*(n+1)*p(n+2) = (14*n+11)*p(n+1) - 3*(2*n+1)*p(n), for n >= 1, with p(0) = 0, p(1) = 3/4, and p(2) = 15/32.

G.f. of p(n): 3*x*(1 - 1/sqrt(4-3*x))/(2*(1-x)). (End)

Assuming offset 0: a(n) = numerator((4^(2*n+1)*n!^2)/((2*n+3)*(2*n)!)). - Peter Luschny, Dec 05 2024