A238616 Number of partitions of n having standard deviation σ < 1.

1, 2, 3, 4, 6, 8, 10, 12, 15, 19, 23, 25, 33, 41, 44, 51, 58, 67, 78, 84, 99, 117, 124, 132, 155, 186, 202, 219, 244, 268, 290, 317, 344, 396, 427, 449, 501, 557, 597, 639, 714, 752, 824, 885, 948, 1031, 1084, 1185, 1308, 1390, 1452, 1589, 1692, 1788, 1919

Offset: 1

Author

Clark Kimberling, Mar 01 2014

Keywords

• nonn
• easy

Comments

Here, "standard deviation" means "population standard deviation" (denoted by σ), not "sample standard deviation" (denoted by s); σ is the square root of variance, so that σ of a list t = (t(k)), such as the partitions of a positive integer, is given by the formula sqrt((Sum_{k=1..#t} (t(k) - mean(t))^2)/(#t)), where #t is the number of terms in t(k). (The distinction between σ and s is discussed in most probability and statistics textbooks. The command "StandardDeviation" in Mathematica gives s, not σ.)

Examples

			There are 11 partitions of 6, whose standard deviations are given by these approximations: 0, 2, 1, 1.41421, 0, 0.816497, 0.866025, 0, 0.5, 0.4, 0, so that a(6) = 8.
		

Crossrefs

Cf. A238617-A238620, A238655-A238662.

Column k=0 of A239223.

Programs

Maple

b:= proc(n, i, m, s, c) `if`(n=0, `if`(s/c-(m/c)^2<1, 1, 0),
      `if`(i=1, b(0$2, m+n, s+n, c+n), add(b(n-i*j, i-1,
       m+i*j, s+i^2*j, c+j), j=0..n/i)))
    end:
a:= n-> b(n$2, 0$3):
seq(a(n), n=1..55);  # Alois P. Heinz, Mar 12 2014

Mathematica

z = 55; g[n_] := g[n] = IntegerPartitions[n]; s[t_] := s[t] = Sqrt[Sum[(t[[k]] - Mean[t])^2, {k, 1, Length[t]}]/Length[t]]
Table[Count[g[n], p_ /; s[p] < 1], {n, z}]   (*A238616*)
Table[Count[g[n], p_ /; s[p] <= 1], {n, z}]  (*A238617*)
Table[Count[g[n], p_ /; s[p] == 1], {n, z}]  (*A238618*)
Table[Count[g[n], p_ /; s[p] > 1], {n, z}]   (*A238619*)
Table[Count[g[n], p_ /; s[p] >= 1], {n, z}]  (*A238620*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]] ListPlot[Sort[t[30]]] (*plot of st.dev's 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<1, 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_] := b[n, n, 0, 0, 0]; Table[a[n], {n, 1, 55}] (* Jean-François Alcover, Nov 20 2015, after Alois P. Heinz *)
(* The interest of this 3rd program is just to show how Mathematica's StandardDeviation can be used, with a correction factor, to compute sigma, the population standard deviation. *)
sigma[t_] := If[Length[t] == 1, 0, StandardDeviation[t]*Sqrt[(Length[t]-1)/ Length[t]]];
a[n_] := Count[IntegerPartitions[n], p_ /; sigma[p] < 1];
Array[a, 30] (* Jean-François Alcover, May 28 2021 *)

Formula

a(n) = A000041(n) - A238620(n).