A238660 Number of partitions of n having population standard deviation = 2.
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
Offset: 1
Examples
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.
Programs
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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(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 = 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_] := b[n, n, 0, 0, 0]; Array[a, 50] (* Jean-François Alcover, Jun 03 2021, after Alois P. Heinz *)
Extensions
a(51)-a(71) from Alois P. Heinz, Mar 11 2014
Comments