|
%I #21 Jun 03 2021 09:31:09
%S 0,0,0,0,1,2,5,8,15,22,33,47,68,93,132,176,239,314,412,536,693,884,
%T 1131,1427,1803,2249,2808,3489,4321,5325,6552,8022,9799,11913,14456,
%U 17502,21136,25457,30588,36673,43869,52398,62437,74277,88186,104526,123670,146028
%N Number of partitions of n having population standard deviation > 1.
%C Regarding "standard deviation" see Comments at A238616.
%F a(n) + A238617(n) = A000041(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) = 2.
%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
|
