A067631 - OEIS

%I #9 Mar 16 2015 22:35:23

%S 0,0,0,0,1,0,0,0,2,0,1,0,4,1,0,0,1,0,2,3,6,0,1,0,8,0,3,0,2,0,0,6,11,1,

%T 1,0,12,7,2,0,3,0,5,1,15,0,0,0,2,10,6,0,1,4,3,11,19,0,1,0,21,2,0,6,5,

%U 0,9,14,3,0,1,0,25,1,10,3,6,0,1,0,28,0,2,8,29,18,5,0,1,4,12,20,32,10,0,0

%N If n is composite then a(n) is the standard deviation of the prime factors of n, rounded off to the nearest integer (rounding up if there's a choice), with each factor counted according to its frequency of occurrence in the prime factorization. If n is 1 or prime then a(n)=0.

%C The (sample) standard deviation sigma of {x_1,...,x_n} is calculated from sigma^2 = 1/(n-1) * sum_{1,...,n}(x_i - mu)^2, where mu denotes the average of {x_1,...,x_n}.

%e 24 = 2^3 * 3^1, so the corresponding average = (2 + 2 + 2 + 3)/ 4 = 2.25 and the standard deviation is [(1/3){3 * (2-2.25)^2 + (3-2.25)^2}]^0.5 = 0.5, which rounds to 1. So a(24) = 1.

%t <<Statistics`NormalDistribution` f[n_] := Flatten[Table[ #[[1]], {#[[2]]}]&/@FactorInteger[n]]; a[n_] := If[PrimeQ[n]||n==1, 0, Floor[StandardDeviation[f[n]]+1/2]]

%K easy,nonn

%O 2,9

%A _Joseph L. Pe_, Feb 02 2002

%E Edited and extended by _Robert G. Wilson v_, Feb 05 2002

%E Edited by _Dean Hickerson_, Feb 12 2002