login
A239140
Number of strict partitions of n having standard deviation σ < 1.
5
1, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1
OFFSET
1,3
COMMENTS
Regarding standard deviation, see Comments at A238616.
FORMULA
a(n + 3) = A083039(n) for n >= 1 (periodic with period 6); a(n) + A239143(n) = A000009(n) for n >=1.
G.f.: -(x^6+x^5+x^4+2*x^3+3*x^2+2*x+1)*x / ((x-1)*(x+1)*(x^2+x+1)). - Alois P. Heinz, Mar 14 2014
EXAMPLE
The standard deviations of the strict partitions of 9 are 0., 3.5, 2.5, 1.5, 2.16025, 0.5, 1.63299, 0.816497, so that a(9) = 3.
MATHEMATICA
z = 30; g[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &]; s[t_] := s[t] = Sqrt[Sum[(t[[k]] - Mean[t])^2, {k, 1, Length[t]}]/Length[t]]
Table[Count[g[n], p_ /; s[p] < 1], {n, z}] (* A239140 *)
Table[Count[g[n], p_ /; s[p] <= 1], {n, z}] (* A239141 *)
Table[Count[g[n], p_ /; s[p] == 1], {n, z}] (* periodic 01 *)
Table[Count[g[n], p_ /; s[p] > 1], {n, z}] (* A239142 *)
Table[Count[g[n], p_ /; s[p] >= 1], {n, z}] (* A239143 *)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsQ[n]}]]
ListPlot[Sort[t[30]]] (*plot of st.dev's of strict partitions of 30*)
(* Peter J. C. Moses, Mar 03 2014 *)
Join[{1, 1, 2}, LinearRecurrence[{-1, 0, 1, 1}, {1, 2, 2, 2}, 83]] (* Ray Chandler, Aug 25 2015 *)
PROG
(PARI)
A083039(n) = (1+!(n%2)+!(n%3));
A239140(n) = if(n<=3, 1+(3==n), A083039(n-3)); \\ Antti Karttunen, May 24 2021
CROSSREFS
Column k=0 of A239228.
Sequence in context: A082641 A334507 A342462 * A138553 A069016 A211270
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Mar 11 2014
EXTENSIONS
A-number in the first formula corrected by Antti Karttunen, May 24 2021
STATUS
approved