A238661 Number of partitions of n having standard deviation σ > 2.

0, 0, 0, 0, 0, 0, 1, 2, 5, 7, 12, 18, 29, 42, 61, 85, 118, 164, 223, 299, 399, 530, 693, 888, 1157, 1488, 1901, 2403, 3044, 3807, 4783, 5935, 7368, 9097, 11197, 13721, 16806, 20441, 24868, 30133, 36494, 43895, 52880, 63424, 75900, 90609, 108088, 128404

Offset: 1

Author

Clark Kimberling, Mar 03 2014

Keywords

• nonn
• easy

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			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.
		

Crossrefs

Cf. A238616, A238658-A238660, A238662.

Programs

Mathematica

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]];
Table[Count[g[n], p_ /; s[p] < 2], {n, z}]   (*A238658*)
Table[Count[g[n], p_ /; s[p] <= 2], {n, z}]  (*A238659*)
Table[Count[g[n], p_ /; s[p] == 2], {n, z}]  (*A238660*)
Table[Count[g[n], p_ /; s[p] > 2], {n, z}]   (*A238661*)
Table[Count[g[n], p_ /; s[p] >= 2], {n, z}]  (*A238662*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
ListPlot[Sort[t[30]]] (*plot of st deviations of partitions of 30*)

Formula

a(n) + A238659(n) = A000041(n).