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%I #18 May 26 2018 22:32:30
%S 1,0,22,210,4426,102330,2906362,95952570,3622138906,153816150810,
%T 7257695358202,376693381614330,21328770664314586,1308295248437904090,
%U 86423208789970618042,6116714829331037666490,461779664078480243085466,37040796099362864616022170
%N Central moments of a Fibonacci-geometric probability distribution.
%C If F(k) is the k-th Fibonacci number, where F(0)=0, F(1)=1, and F(n)=F(n-1)+F(n-2), then p(k)=F(k-1)/2^k is a normalized probability distribution on the positive integers.
%C For example, it is the probability that k coin tosses are required to get two heads in a row, or the probability that a random series of k bits has its first two consecutive 1's at the end.
%C The g.f. for this distribution is g(x) = x^2/(4-2x-x^2) = (1/4)x^2 + (1/8)x^3 + (1/8)x^4 + (3/32)x^5 + ....
%C The mean of this distribution is 6. (See A302922.)
%C The n-th moments about the mean, known as central moments, are defined by a(n) = Sum_{k>=1} ((k-6)^n)p(k). They appear to be integers and form this sequence.
%C For n >= 1, a(n) appears to be even. Dividing these terms by 2 gives sequence A302925.
%C The raw moments (i.e., the moments about zero) also appear to be integers. This is sequence A302922.
%C The raw moments also appear to be even for n >= 1. Dividing them by 2 gives sequence A302923.
%C The cumulants of this distribution, defined by the cumulant e.g.f. log(g(e^x)), also appear to be integers. They form sequence A302926.
%C The cumulants also appear to be even for n >= 0. Dividing them by 2 gives sequence A302927.
%C Note: Another probability distribution on the positive integers that has integral moments and cumulants is the geometric distribution p(k)=1/2^k. The sequences related to these moments are A000629, A000670, A052841, and A091346.
%H Albert Gordon Smith, <a href="/A302924/b302924.txt">Table of n, a(n) for n = 0..300</a>
%H Christopher Genovese, <a href="http://www.stat.cmu.edu/~genovese/class/iprob-S06/notes/double-heads.pdf">Double Heads</a>
%F In the following,
%F F(k) is the k-th Fibonacci number, as defined in the Comments.
%F phi=(1+sqrt(5))/2 is the golden ratio, and psi=(1-sqrt(5))/2.
%F LerchPhi(z,s,a) = Sum_{k>=0} z^k/(a+k)^s is the Lerch transcendant.
%F For n >= 0:
%F a(n) = Sum_{k>=1} (((k-6)^n)(F(k-1)/2^k));
%F a(n) = Sum_{k>=1} (((k-6)^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));
%F a(n) = (LerchPhi(phi/2,-n,-5)-LerchPhi(psi/2,-n,-5))/(2 sqrt(5));
%F a(n) = Sum_{k=0..n} (binomial(n,k)*A302922(k)*(-6)^(n-k)).
%e a(0)=1 is the 0th central moment of the distribution, which is the total probability.
%e a(1)=0 is the 1st central moment, or the "mean about the mean". It is zero by definition of central moments.
%e a(2)=22 is the 2nd central moment, known as the variance or the square of the standard deviation. It measures how far integers following the distribution are from the mean by averaging the squares of their differences from the mean.
%t Module[{max, r, g, moments},
%t max = 17;
%t r = Range[0, max];
%t g[x_] := x^2/(4 - 2 x - x^2);
%t moments = r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x];
%t Table[Sum[Binomial[n, k] moments[[k + 1]] (-6)^(n - k), {k, 0, n}], {n, 0, max}]
%t ]
%Y Central half-moments: A302925.
%Y Raw moments: A302922.
%Y Raw half-moments: A302923.
%Y Cumulants: A302926.
%Y Half-cumulants: A302927.
%Y Cf. A000629, A000670, A052841, A091346.
%K nonn
%O 0,3
%A _Albert Gordon Smith_, Apr 15 2018
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