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%I #24 Mar 26 2018 04:24:44
%S 0,0,1,4,47,244,1186,1384,25147,112028,98374,1067720,1531401,39249768,
%T 166656772,88008656,2961699667,12412521388,51854046982,108006842264,
%U 448816369361,3721813363288,15401045060572,15904199160592,131178778841711,1080387930269464,4443100381114156,9124976352166288
%N Numerator of variance of n-th row of Pascal's triangle.
%C Variance here is the sample variance unbiased estimator. For population variance, see A301631.
%H Chai Wah Wu, <a href="/A301278/b301278.txt">Table of n, a(n) for n = 0..1659</a>
%H Simon Demers, <a href="https://doi.org/10.1080/00031305.2017.1422439">Taylor's Law Holds for Finite OEIS Integer Sequences and Binomial Coefficients</a>, American Statistician, online: 19 Jan 2018.
%F a(0) = 0; a(n) = numerator of binomial(2n,n)/n - 4^n/(n*(n+1)) for n >= 1. - _Chai Wah Wu_, Mar 23 2018
%e The first few variances are 0, 0, 1/3, 4/3, 47/10, 244/15, 1186/21, 1384/7, 25147/36, 112028/45, 98374/11, 1067720/33, 1531401/13, 39249768/91, 166656772/105, 88008656/15, 2961699667/136, 12412521388/153, 51854046982/171, 108006842264/95, 448816369361/105, ...
%p M:=70;
%p m := n -> 2^n/(n+1);
%p m1:=[seq(m(n),n=0..M)]; # A084623/A000265
%p v := n -> (1/n) * add((binomial(n,i) - m(n))^2, i=0..n );
%p v1:= [0, 0, seq(v(n),n=2..60)]; # A301278/A301279
%o (Python)
%o from fractions import Fraction
%o from sympy import binomial
%o def A301278(n):
%o return (Fraction(int(binomial(2*n,n)))/n - Fraction(4**n)/(n*(n+1))).numerator if n > 0 else 0 # _Chai Wah Wu_, Mar 23 2018
%o (PARI) a(n) = if(n==0, 0, numerator(binomial(2*n,n)/n - 4^n/(n*(n+1)))); \\ _Altug Alkan_, Mar 25 2018
%Y Mean and variance of n-th row of Pascal's triangle: A084623/A000265, A301278/A301279, A054650, A301280.
%K nonn,frac
%O 0,4
%A _N. J. A. Sloane_, Mar 18 2018
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