A238618 Number of partitions of n having standard deviation σ = 1.

0, 0, 0, 1, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 4, 0, 4, 0, 7, 0, 1, 0, 16, 0, 1, 0, 10, 0, 11, 0, 10, 0, 1, 0, 26, 0, 1, 0, 26, 0, 24, 0, 13, 0, 1, 0, 60, 0, 5, 0, 17, 0, 19, 0, 52, 0, 1, 0, 117, 0, 1, 0, 36, 0, 46, 0, 23, 0, 29, 0, 160, 0, 1, 0, 30, 0, 61, 0, 140

Offset

1,8

Comments

Regarding "standard deviation" see Comments at A238616.

Links

Table of n, a(n) for n=1..80.

Example

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) = 1.

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..50);  # Alois P. Heinz, Mar 11 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*)
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, 50}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)

Crossrefs

Cf. A238616.

Sequence in context: A362837 A276193 A357400 * A132277 A137286 A180048

Adjacent sequences: A238615 A238616 A238617 * A238619 A238620 A238621

Keyword

nonn,easy

Author

Clark Kimberling, Mar 01 2014

Extensions

a(56)-a(80) from Alois P. Heinz, Mar 11 2014

Status

approved