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%I #16 Mar 11 2014 13:38:48
%S 0,0,0,0,0,0,0,0,1,2,5,5,10,15,23,33,49,68,99,133,179,246,334,432,583,
%T 756,974,1261,1618,2076,2657,3336,4228,5270,6592,8190,10182,12567,
%U 15533,19008,23307,28410,34622,42041,50959,61487,74259,88734,106666,127587
%N Number of partitions of n having standard deviation σ > 3.
%C Regarding "standard deviation" see Comments at A238616.
%e There are 30 partitions of 9, whose standard deviations are given by these approximations: 0., 3.5, 2.5, 2.82843, 1.5, 2.16025, 2.16506, 0.5, 1.63299, 1.41421, 1.63936, 1.6, 1.41421, 0.816497, 1.29904, 1.08972, 1.16619, 1.11803, 0., 0.829156, 0.979796, 0.433013, 0.748331, 0.763763, 0.699854, 0.4, 0.5, 0.451754, 0.330719, 0, so that a(9) = 1.
%t z = 53; g[n_] := g[n] = IntegerPartitions[n]; c[t_] := c[t] = Length[t];
%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] > 3], {n, z}] (*A238655*)
%t Table[Count[g[n], p_ /; s[p] > 4], {n, z}] (*A238656*)
%t Table[Count[g[n], p_ /; s[p] > 5], {n, z}] (*A238657*)
%t t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
%t ListPlot[Sort[t[30]]] (*plot of st dev's of partitions of 30*)
%Y Cf. A238616, A238661, A238656, A238657.
%K nonn,easy
%O 1,10
%A _Clark Kimberling_, Mar 03 2014
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