A157683 Decimal expansion of Levy-Cantor distribution area constant.

6, 8, 8, 5, 4, 9, 5, 7, 8, 7, 0, 2, 7, 2, 6, 1, 8, 6, 9, 6, 3, 1, 9, 5, 5, 2, 1, 5, 5, 5, 7, 7, 6, 7, 6, 1, 1, 7, 4, 8, 1, 3, 9, 4, 6, 6, 5, 6, 8, 8, 0, 4, 4, 1, 7, 9, 4, 4, 9, 3, 0, 9, 9, 4, 6, 2, 7, 5, 4, 3, 8, 7, 8, 9, 0, 5, 3, 6, 7, 3, 9, 4, 2, 9, 5, 6, 8, 0, 3, 2, 1, 1, 3, 8, 4, 3, 2, 6, 1, 8

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

Sequence in context: A011481 A212480 A100221 * A300897 A346289 A049110

Adjacent sequences: A157680 A157681 A157682 * A157684 A157685 A157686

Keyword

nonn,cons

Author

Roger L. Bagula, Mar 04 2009

Extensions

Edited by Charles R Greathouse IV, Feb 03 2011

Status

approved