A238659 Number of partitions of n having standard deviation σ <= 2.

1, 2, 3, 5, 7, 11, 14, 20, 25, 35, 44, 59, 72, 93, 115, 146, 179, 221, 267, 328, 393, 472, 562, 687, 801, 948, 1109, 1315, 1521, 1797, 2059, 2414, 2775, 3213, 3686, 4256, 4831, 5574, 6317, 7205, 8089, 9279, 10381, 11751, 13234, 14949, 16666, 18869, 20986

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) = 20.
		

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) + A238661(n) = A000041(n).