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%I #14 Mar 11 2014 13:40:19
%S 0,0,0,0,0,0,1,2,5,7,12,18,29,42,61,85,118,164,223,299,399,530,693,
%T 888,1157,1488,1901,2403,3044,3807,4783,5935,7368,9097,11197,13721,
%U 16806,20441,24868,30133,36494,43895,52880,63424,75900,90609,108088,128404
%N Number of partitions of n having standard deviation σ > 2.
%C Regarding "standard deviation" see Comments at A238616.
%F a(n) + A238659(n) = A000041(n).
%e There are 22 partitions of 8, whose standard deviations are given by these approximations: 0., 3., 2., 2.35702, 1., 1.69967, 1.73205, 0., 1.24722, 0.942809, 1.22474, 1.2, 0.471405, 1., 0.707107, 0.8, 0.745356, 0., 0.489898, 0.471405, 0.349927, 0, so that a(8) = 2.
%t z = 50; g[n_] := g[n] = IntegerPartitions[n]; c[t_] := c[t] = Length[t]; s[t_] := s[t] = Sqrt[Sum[(t[[k]] - Mean[t])^2, {k, 1, c[t]}]/c[t]];
%t Table[Count[g[n], p_ /; s[p] < 2], {n, z}] (*A238658*)
%t Table[Count[g[n], p_ /; s[p] <= 2], {n, z}] (*A238659*)
%t Table[Count[g[n], p_ /; s[p] == 2], {n, z}] (*A238660*)
%t Table[Count[g[n], p_ /; s[p] > 2], {n, z}] (*A238661*)
%t Table[Count[g[n], p_ /; s[p] >= 2], {n, z}] (*A238662*)
%t t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
%t ListPlot[Sort[t[30]]] (*plot of st deviations of partitions of 30*)
%Y Cf. A238616, A238658-A238660, A238662.
%K nonn,easy
%O 1,8
%A _Clark Kimberling_, Mar 03 2014