A238657 Number of partitions of n having standard deviation σ > 5.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 5, 9, 11, 16, 25, 34, 45, 64, 87, 121, 160, 212, 279, 369, 481, 614, 797, 1027, 1308, 1670, 2102, 2661, 3345, 4189, 5224, 6494, 8069, 9982, 12281, 15093, 18508, 22731, 27564, 33639, 40757, 49496, 59838, 72228

Offset

1,14

Comments

Regarding "standard deviation" see Comments at A238616.

Links

Table of n, a(n) for n=1..53.

Example

There are 30 partitions of 9, whose standard deviations are given by these approximations:  0., 3.5, 2.5, 2.82843, 1.5, 2.16025, 2.16506, 0.5, 1.63299, 1.41421, 1.63936, 1.6, 1.41421, 0.816497, 1.29904, 1.08972, 1.16619, 1.11803, 0., 0.829156, 0.979796, 0.433013, 0.748331, 0.763763, 0.699854, 0.4, 0.5, 0.451754, 0.330719, 0, so that a(9) = 0.

Mathematica

z = 53; 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] > 3], {n, z}]   (*A238655*)
Table[Count[g[n], p_ /; s[p] > 4], {n, z}]   (*A238656*)
Table[Count[g[n], p_ /; s[p] > 5], {n, z}]   (*A238657*)
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*)

Crossrefs

Cf. A238616, A238661, A238655, A238657.

Sequence in context: A324698 A059042 A157604 * A329327 A329357 A101737

Adjacent sequences: A238654 A238655 A238656 * A238658 A238659 A238660

Keyword

nonn,easy

Author

Clark Kimberling, Mar 03 2014

Status

approved