A238658 Number of partitions of n having population standard deviation < 2.

1, 2, 3, 5, 7, 10, 14, 19, 25, 33, 44, 57, 72, 92, 114, 143, 179, 216, 267, 321, 389, 470, 562, 668, 798, 946, 1100, 1295, 1521, 1759, 2059, 2392, 2742, 3206, 3674, 4172, 4831, 5566, 6265, 7115, 8089, 9152, 10381, 11664, 13131, 14927, 16666, 18565, 20977

Offset: 1

Author

Clark Kimberling, Mar 03 2014

Keywords

• nonn
• easy

Examples

			There are 22 partitions of 8, whose population 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) = 19.
		

Crossrefs

Cf. A000041, A238616, A238659, A238660, A238661, 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 *)
(* Second program: *)
b[n_, i_, m_, s_, c_] := b[n, i, m, s, c] = If[n == 0, If[s/c - (m/c)^2 >= 4, 1, 0], If[i == 1, b[0, 0, m + n, s + n, c + n], Sum[b[n - i*j, i - 1, m + i*j, s + i^2*j, c + j], {j, 0, n/i}]]];
a[n_] := PartitionsP[n] - b[n, n, 0, 0, 0];
Array[a, 50] (* Jean-François Alcover, May 27 2021, after Alois P. Heinz *)

Formula

a(n) + A238662(n) = A000041(n).