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A238655 Number of partitions of n having standard deviation σ > 3.

%I A238655 #16 Mar 11 2014 13:38:48
%S A238655 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 A238655 756,974,1261,1618,2076,2657,3336,4228,5270,6592,8190,10182,12567,
%U A238655 15533,19008,23307,28410,34622,42041,50959,61487,74259,88734,106666,127587
%N A238655 Number of partitions of n having standard deviation σ > 3.
%C A238655 Regarding "standard deviation" see Comments at A238616.
%e A238655 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 A238655 z = 53; g[n_] := g[n] = IntegerPartitions[n]; c[t_] := c[t] = Length[t];
%t A238655 s[t_] := s[t] = Sqrt[Sum[(t[[k]] - Mean[t])^2, {k, 1, c[t]}]/c[t]]
%t A238655 Table[Count[g[n], p_ /; s[p] > 3], {n, z}]   (*A238655*)
%t A238655 Table[Count[g[n], p_ /; s[p] > 4], {n, z}]   (*A238656*)
%t A238655 Table[Count[g[n], p_ /; s[p] > 5], {n, z}]   (*A238657*)
%t A238655 t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
%t A238655 ListPlot[Sort[t[30]]] (*plot of st dev's of partitions of 30*)
%Y A238655 Cf. A238616, A238661, A238656, A238657.
%K A238655 nonn,easy
%O A238655 1,10
%A A238655 _Clark Kimberling_, Mar 03 2014