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A395207
Numerator of the probability that the unique circle passing through three points, selected independently and uniformly at random in an n-ball, is entirely contained within that n-ball, divided by Pi^2 if n is odd.
8
2, 12, 14, 600, 11, 7840, 494, 105840, 37145, 8964648, 392863, 64128064, 131905, 2473511040, 117285338, 56166368544, 3650109911, 426684744200, 860010277, 46593974066640, 584803025179, 551568386713344, 39181802686993, 36562298361680000, 103559043001879, 3153558501219672000
OFFSET
2,1
COMMENTS
The probability that the three points are collinear (i.e., lie on the same straight line) and do not define a circle is 0.
LINKS
Fernando Affentranger, Random circles in the d-dimensional unit ball, Journal of Applied Probability, Vol. 26, No. 2 (1989), p. 408-412; JSTOR link.
FORMULA
Let f(n) = a(n)/A395208(n) if n is even, and a(n)*Pi^2/A395208(n) if n is odd. Then:
f(n) = Pi * (n-1) * n^2 * Gamma(n/2)^3 * Gamma((3*n-1)/2) / (6 * (2*n+1) * Gamma((n+1)/2)^3 * Gamma(3*n/2)).
Limit_{n->oo} f(n) = Pi/(3*sqrt(3)) = 0.604599... (A073010).
f(n) ~ Pi/(3*sqrt(3))*(1 - 1/(2*n) - 7/(36*n^2) - 25/(72*n^3) + 11/(432*n^4) + ...).
EXAMPLE
Fractions begin with 2/5, 12*Pi^2/245, 14/27, 600*Pi^2/11011, 11/20, 7840*Pi^2/138567, 494/875, 105840*Pi^2/1834963, 37145/64827, 8964648*Pi^2/153609433, 392863/679140, 64128064*Pi^2/1090194825, ...
MATHEMATICA
a[n_] := Numerator[Pi*(n-1) * n^2 * Gamma[n/2]^3 * Gamma[(3*n-1)/2] / (6 *(2*n+1) * Gamma[(n+1)/2]^3 * Gamma[3*n/2])] /. Pi -> 1; Array[a, 26, 2]
PROG
(PARI) gamma_my(n) = if(denominator(n) == 1, (n-1)!, my(m = n - 1/2); (2*m)! / (4^m * m!)); \\ gamma for integers, gamma(n)/sqrt(Pi) for half-integers
a(n) = numerator((n-1) * n^2 * gamma_my(n/2)^3 * gamma_my((3*n-1)/2) / (6 * (2*n+1) * gamma_my((n+1)/2)^3 * gamma_my(3*n/2)));
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Amiram Eldar, Apr 16 2026
STATUS
approved