OFFSET
1,1
COMMENTS
The idea that the Levy distribution could be Bernoulli-like gave me this result:
m = mean, a = Fractal dimension as the Levy exponent,
f(x, a, m) = (1 + a) * log(x - m)/((x - m)^(1 + a) - 1)
On a {0,1} domain with mean = 1/2 you have to use the absolute value function.
It gives a cusp or Lambda point at 1/2.
Here is what it came from:
f(x, m, t) = t/((x-m)^(1+a) + t^(1+a)): Levy distribution;
as Bernoulli in t:
f(t) = t/(exp(t) - 1), so exp(t) = exp(log[(x-m)^(1+a) + t^(1+a) + 1)), and it gives t = (1+a)*log(x-m) + log(1 + (t^(1+a) + 1)/(x-m)^(1+a))
Limit for large (x-m)^(1+a) is: t = (1+a)*log(x-m).
In the Bernoulli f(t) that gives the distribution. Cantor dimension is a = log(2)/log(3). So taking it with m = 0, you get a finite constant area of a long tailed Levy-Cantor distribution that is Bernoulli-Boson like.
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..10000
EXAMPLE
6.88549578702726186963195521555776761174813946656880441794493...
MATHEMATICA
a = Log[2]/Log[3]; m = 0; c = Integrate[(1 + a) * Log[x - m]/((x - m)^(1 + a) - 1), {x, 0, Infinity}] (* Bagula *)
RealDigits[Log[6, 3]*(PolyGamma[1, Log[6, 2]] + PolyGamma[1, Log[6, 3]]), 10, 100][[1]] (* Charles R Greathouse IV, Feb 03 2011 *)
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Roger L. Bagula, Mar 04 2009
EXTENSIONS
Edited by Charles R Greathouse IV, Feb 03 2011
STATUS
approved