A238620 - OEIS

%I #22 May 12 2023 12:16:53

%S 0,0,0,1,1,3,5,10,15,23,33,52,68,94,132,180,239,318,412,543,693,885,

%T 1131,1443,1803,2250,2808,3499,4321,5336,6552,8032,9799,11914,14456,

%U 17528,21136,25458,30588,36699,43869,52422,62437,74290,88186,104527,123670

%N Number of partitions of n having population standard deviation >= 1.

%C Regarding "standard deviation" see Comments at A238616.

%F a(n) = A000041(n) - A238616(n).

%e 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) = 3.

%p b:= proc(n, i, m, s, c) `if`(n=0, `if`(s/c-(m/c)^2>=1, 1, 0),

%p       `if`(i=1, b(0$2, m+n, s+n, c+n), add(b(n-i*j, i-1,

%p        m+i*j, s+i^2*j, c+j), j=0..n/i)))

%p     end:

%p a:= n-> b(n$2, 0$3):

%p seq(a(n), n=1..50);  # _Alois P. Heinz_, Mar 11 2014

%t 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]]

%t Table[Count[g[n], p_ /; s[p] < 1], {n, z}]   (*A238616*)

%t Table[Count[g[n], p_ /; s[p] <= 1], {n, z}]  (*A238617*)

%t Table[Count[g[n], p_ /; s[p] == 1], {n, z}]  (*A238618*)

%t Table[Count[g[n], p_ /; s[p] > 1], {n, z}]   (*A238619*)

%t Table[Count[g[n], p_ /; s[p] >= 1], {n, z}]  (*A238620*)

%t t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]

%t ListPlot[Sort[t[30]]] (*plot of st.dev's of partitions of 30*)

%t (* Second program: *)

%t b[n_, i_, m_, s_, c_] := b[n, i, m, s, c] = If[n == 0,

%t      If[s/c - (m/c)^2 >= 1, 1, 0], If[i == 1, b[0, 0, m+n, s+n, c+n],

%t      Sum[b[n - i*j, i - 1, m + i*j, s + i^2*j, c + j], {j, 0, n/i}]]];

%t a[n_] := b[n, n, 0, 0, 0];

%t Array[a, 50] (* _Jean-François Alcover_, Jun 03 2021, after _Alois P. Heinz_ *)

%Y Cf. A238616.

%K nonn,easy

%O 1,6

%A _Clark Kimberling_, Mar 01 2014