A302925 - OEIS

%I #21 May 26 2018 22:33:18

%S 0,11,105,2213,51165,1453181,47976285,1811069453,76908075405,

%T 3628847679101,188346690807165,10664385332157293,654147624218952045,

%U 43211604394985309021,3058357414665518833245,230889832039240121542733,18520398049681432308011085

%N Central 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 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 A302924.

%C For n >= 1, that sequence appears to be even. Dividing those terms by 2 gives this sequence.

%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 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="/A302925/b302925.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 LerchPhi(z,s,a) = Sum_{k>=0} z^k/(a+k)^s is the Lerch transcendant.

%F For n>=1:

%F a(n) = (1/2)A302924(n);

%F a(n) = (1/2)Sum_{k>=1} (((k-6)^n)(F(k-1)/2^k));

%F a(n) = (1/2)Sum_{k>=1} (((k-6)^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));

%F a(n) = (1/2)(LerchPhi(phi/2,-n,-5)-LerchPhi(psi/2,-n,-5))/(2 sqrt(5));

%F a(n) = (1/2)Sum_{k=0..n} (binomial(n,k)A302922(k)(-6)^(n-k)).

%e a(1)=0 is half the 1st central moment of the distribution, or half the "mean about the mean". It is zero by definition of central moments.

%e a(2)=11 is half the 2nd central moment, or half the variance, or half the square of the standard deviation.

%t Module[{max, r, g, moments, centralMoments},

%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   centralMoments = Table[Sum[Binomial[n, k] moments[[k + 1]] (-6)^(n - k), {k, 0, n}], {n, 0, max}];

%t   Rest[centralMoments]/2

%t ]

%Y Central moments: A302924.

%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 1,2

%A _Albert Gordon Smith_, Apr 15 2018