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%I #29 Apr 13 2022 12:38:09
%S 1,6,58,822,15514,366006,10361818,342239862,12918651034,548600581686,
%T 25885279045978,1343513774912502,76071145660848154,
%U 4666162902628259766,308236822886732856538,21815861409181135034742,1646982315540717414270874,132109620398598537723816246
%N Raw 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 this sequence.
%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, a(n) appears to be even.
%C Dividing these terms by 2 gives sequence A302923.
%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, and A091346.
%H Albert Gordon Smith, <a href="/A302922/b302922.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 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>=0:
%F a(n) = Sum_{k>=1} ((k^n)(F(k-1)/2^k));
%F a(n) = Sum_{k>=1} ((k^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));
%F a(n) = (Li(-n,phi/2)/phi-Li(-n,psi/2)/psi)/sqrt(5).
%F E.g.f.: g(e^x) where g(x) = x^2/(4-2x-x^2) is the g.f. for the probability distribution.
%F a(n) ~ n! * (5 - sqrt(5)) / (10 * (log(sqrt(5) - 1))^(n+1)). - _Vaclav Kotesovec_, Apr 13 2022
%e a(0)=1 is the 0th raw moment of the distribution, which is the total probability.
%e a(1)=6 is the 1st raw moment, known as the mean of the distribution. It is the arithmetic average of integers following the distribution.
%e a(2)=58 is the 2nd raw moment. It is the arithmetic average of the squares of integers following the distribution.
%t Module[{max, r, g},
%t max = 17;
%t r = Range[0, max];
%t g[x_] := x^2/(4 - 2 x - x^2);
%t r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x]
%t ]
%o (PARI) Vec(serlaplace(exp(2*x)/(4-2*exp(x)-exp(2*x)))) \\ _Michel Marcus_, Apr 17 2018
%Y Raw half-moments: A302923.
%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 0,2
%A _Albert Gordon Smith_, Apr 15 2018
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