A302927 Half-cumulants of a Fibonacci-geometric probability distribution.
0, 3, 11, 105, 1487, 28065, 662111, 18744705, 619117727, 23370059265, 992427775391, 46826909698305, 2430439250993567, 137613995282046465, 8441167989376455071, 557605650894240475905, 39465264036137261935007, 2979418998248159878129665
Offset: 0
Keywords
Examples
a(0)=0 is half the 0th cumulant of the distribution. The 0th cumulant is always zero. a(1)=3 is half the 1st cumulant, which is half the mean. a(2)=11 is half the 2nd cumulant, which is half the variance.
Links
- Albert Gordon Smith, Table of n, a(n) for n = 0..300
- Christopher Genovese, Double Heads
Crossrefs
Programs
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Mathematica
Module[{max, r, g}, max = 17; r = Range[0, max]; g[x_] := x^2/(4 - 2 x - x^2); (1/2) r! CoefficientList[Normal[Series[Log[g[Exp[x]]], {x, 0, max}]], x] ]
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PARI
concat(0, Vec(serlaplace(log(exp(2*x)/(4-2*exp(x)-exp(2*x))))/2)) \\ Michel Marcus, Apr 17 2018
Formula
E.g.f.: (1/2)*log(g(e^x)) where g(x) = x^2/(4-2*x-x^2) is the g.f. for the probability distribution.
Comments