%I #26 May 26 2018 22:32:04
%S 3,29,411,7757,183003,5180909,171119931,6459325517,274300290843,
%T 12942639522989,671756887456251,38035572830424077,2333081451314129883,
%U 154118411443366428269,10907930704590567517371,823491157770358707135437,66054810199299268861908123
%N Raw half-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 n-th moments about zero of this distribution, known as raw moments, are defined by a(n) = Sum_{k>=1} (k^n)*p(k). They appear to be integers and form A302922.
%C The e.g.f. for the raw moments is g(e^x) = 1 + 6x + 58x^2/2! + 822x^3/3! + ....
%C For n >= 1, the raw moments appear to be even. Dividing them by 2 gives this sequence of raw half-moments.
%C The central moments (i.e., the moments about the mean) also appear to be integers. They form sequence A302924.
%C The central moments also appear to be even for n >= 1. Dividing them by 2 gives sequence A302925.
%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, A091346.
%H Albert Gordon Smith, <a href="/A302923/b302923.txt">Table of n, a(n) for n = 1..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 Li(s,z) is the polylogarithm of order s and argument z.
%F When s is a negative integer as it is here, Li(s,z) is a rational function of z: Li(-n,z) = (z(d/dz))^n(z/(1-z)).
%F For n>=1:
%F a(n) = (1/2)A302922(n);
%F a(n) = (1/2)Sum_{k>=1} ((k^n)(F(k-1)/2^k));
%F a(n) = (1/2)Sum_{k>=1} ((k^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));
%F a(n) = (1/2)(Li(-n,phi/2)/phi-Li(-n,psi/2)/psi)/sqrt(5).
%F E.g.f.: (1/2)g(e^x) where g(x) = x^2/(4-2x-x^2) is the g.f. for the probability distribution.
%e a(1)=3 is half the first raw moment of the distribution. It is half the arithmetic average of integers following the distribution.
%e a(2)=29 is half the second raw moment. It is half the arithmetic average of the squares of integers following the distribution.
%t Module[{max, r, g, rawMoments},
%t max = 17;
%t r = Range[0, max];
%t g[x_] := x^2/(4 - 2 x - x^2);
%t rawMoments = r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x];
%t Rest[rawMoments]/2
%t ]
%Y Raw moments: A302922.
%Y Central moments: A302924.
%Y Central half-moments: A302925.
%Y Cumulants: A302926.
%Y Half-cumulants: A302927.
%Y Cf. A000629, A000670, A052841, A091346.
%K nonn
%O 1,1
%A _Albert Gordon Smith_, Apr 15 2018