

A238660


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


5



0, 0, 0, 0, 0, 1, 0, 1, 0, 2, 0, 2, 0, 1, 1, 3, 0, 5, 0, 7, 4, 2, 0, 19, 3, 2, 9, 20, 0, 38, 0, 22, 33, 7, 12, 84, 0, 8, 52, 90, 0, 127, 0, 87, 103, 22, 0, 304, 9, 74, 131, 153, 0, 214, 139, 390, 192, 59, 0, 1219, 0, 73, 460, 372, 383, 908, 0, 501, 439, 832, 0
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OFFSET

1,10


COMMENTS

Regarding "standard deviation" see Comments at A238616.


LINKS

Table of n, a(n) for n=1..71.


EXAMPLE

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) = 1.


MAPLE

b:= proc(n, i, m, s, c) `if`(n=0, `if`(s/c(m/c)^2=4, 1, 0),
`if`(i=1, b(0$2, m+n, s+n, c+n), add(b(ni*j, i1,
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 = 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*)


CROSSREFS

Cf. A238616, A238658A238660, A238662.
Sequence in context: A279048 A263485 A263489 * A174793 A245718 A192011
Adjacent sequences: A238657 A238658 A238659 * A238661 A238662 A238663


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Mar 03 2014


EXTENSIONS

a(51)a(71) from Alois P. Heinz, Mar 11 2014


STATUS

approved



