A302925 Central half-moments of a Fibonacci-geometric probability distribution.

0, 11, 105, 2213, 51165, 1453181, 47976285, 1811069453, 76908075405, 3628847679101, 188346690807165, 10664385332157293, 654147624218952045, 43211604394985309021, 3058357414665518833245, 230889832039240121542733, 18520398049681432308011085

Offset: 1

Author

Albert Gordon Smith, Apr 15 2018

Keywords

• nonn

Comments

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.

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.

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 + ....

The mean of this distribution is 6. (See A302922.)

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.

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

The raw moments (i.e., the moments about zero) also appear to be integers. This is sequence A302922.

The raw moments also appear to be even for n >= 1. Dividing them by 2 gives sequence A302923.

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.

The cumulants also appear to be even for n >= 0. Dividing them by 2 gives sequence A302927.

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.

Examples

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

Links

• Albert Gordon Smith, Table of n, a(n) for n = 1..300
• Christopher Genovese, Double Heads

Crossrefs

Central moments: A302924.

Raw moments: A302922.

Raw half-moments: A302923.

Cumulants: A302926.

Half-cumulants: A302927.

Cf. A000629, A000670, A052841, A091346.

Programs

Mathematica

Module[{max, r, g, moments, centralMoments},
  max = 17;
  r = Range[0, max];
  g[x_] := x^2/(4 - 2 x - x^2);
  moments = r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x];
  centralMoments = Table[Sum[Binomial[n, k] moments[[k + 1]] (-6)^(n - k), {k, 0, n}], {n, 0, max}];
  Rest[centralMoments]/2
]

Formula

In the following,

F(k) is the k-th Fibonacci number, as defined in the Comments.

phi=(1+sqrt(5))/2 is the golden ratio, and psi=(1-sqrt(5))/2.

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

For n>=1:

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

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

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

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

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