cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 17 results. Next

A238662 Number of partitions of n having population standard deviation >= 2.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 3, 5, 9, 12, 20, 29, 43, 62, 88, 118, 169, 223, 306, 403, 532, 693, 907, 1160, 1490, 1910, 2423, 3044, 3845, 4783, 5957, 7401, 9104, 11209, 13805, 16806, 20449, 24920, 30223, 36494, 44022, 52880, 63511, 76003, 90631, 108088, 128708
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Comments

Regarding "population standard deviation" see Comments at A238616.

Examples

			There are 22 partitions of 8, whose population 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) = 3.

Crossrefs

Cf. A238616, A238658, A238660, A238661.

Programs

  • 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, May 27 2021, after Alois P. Heinz *)

Formula

a(n) + A238658(n) = A000041(n).

A238619 Number of partitions of n having population standard deviation > 1.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 5, 8, 15, 22, 33, 47, 68, 93, 132, 176, 239, 314, 412, 536, 693, 884, 1131, 1427, 1803, 2249, 2808, 3489, 4321, 5325, 6552, 8022, 9799, 11913, 14456, 17502, 21136, 25457, 30588, 36673, 43869, 52398, 62437, 74277, 88186, 104526, 123670, 146028
Offset: 1

Views

Author

Clark Kimberling, Mar 01 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			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.

Crossrefs

Cf. A238616.

Programs

  • Maple
    b:= proc(n, i, m, s, c) `if`(n=0, `if`(s/c-(m/c)^2>1, 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 = 55; g[n_] := g[n] = IntegerPartitions[n]; 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}]   (*A238616*)
Table[Count[g[n], p_ /; s[p] <= 1], {n, z}]  (*A238617*)
Table[Count[g[n], p_ /; s[p] == 1], {n, z}]  (*A238618*)
Table[Count[g[n], p_ /; s[p] > 1], {n, z}]   (*A238619*)
Table[Count[g[n], p_ /; s[p] >= 1], {n, z}]  (*A238620*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
ListPlot[Sort[t[30]]] (*plot of st.dev's 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 > 1, 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 *)

Formula

a(n) + A238617(n) = A000041(n).

A238620 Number of partitions of n having population standard deviation >= 1.

Original entry on oeis.org

0, 0, 0, 1, 1, 3, 5, 10, 15, 23, 33, 52, 68, 94, 132, 180, 239, 318, 412, 543, 693, 885, 1131, 1443, 1803, 2250, 2808, 3499, 4321, 5336, 6552, 8032, 9799, 11914, 14456, 17528, 21136, 25458, 30588, 36699, 43869, 52422, 62437, 74290, 88186, 104527, 123670
Offset: 1

Views

Author

Clark Kimberling, Mar 01 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			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.

Crossrefs

Cf. A238616.

Programs

  • Maple
    b:= proc(n, i, m, s, c) `if`(n=0, `if`(s/c-(m/c)^2>=1, 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 = 55; g[n_] := g[n] = IntegerPartitions[n]; 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}]   (*A238616*)
Table[Count[g[n], p_ /; s[p] <= 1], {n, z}]  (*A238617*)
Table[Count[g[n], p_ /; s[p] == 1], {n, z}]  (*A238618*)
Table[Count[g[n], p_ /; s[p] > 1], {n, z}]   (*A238619*)
Table[Count[g[n], p_ /; s[p] >= 1], {n, z}]  (*A238620*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
ListPlot[Sort[t[30]]] (*plot of st.dev's 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 >= 1, 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 *)

Formula

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

A238658 Number of partitions of n having population standard deviation < 2.

Original entry on oeis.org

1, 2, 3, 5, 7, 10, 14, 19, 25, 33, 44, 57, 72, 92, 114, 143, 179, 216, 267, 321, 389, 470, 562, 668, 798, 946, 1100, 1295, 1521, 1759, 2059, 2392, 2742, 3206, 3674, 4172, 4831, 5566, 6265, 7115, 8089, 9152, 10381, 11664, 13131, 14927, 16666, 18565, 20977
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Examples

			There are 22 partitions of 8, whose population 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) = 19.

Crossrefs

Cf. A000041, A238616, A238659, A238660, A238661, A238662.

Programs

  • 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_] := PartitionsP[n] - b[n, n, 0, 0, 0];
Array[a, 50] (* Jean-François Alcover, May 27 2021, after Alois P. Heinz *)

Formula

a(n) + A238662(n) = A000041(n).

A238660 Number of partitions of n having population standard deviation = 2.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

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.

Crossrefs

Cf. A238616, A238658-A238660, A238662.

Programs

  • 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

A238661 Number of partitions of n having standard deviation σ > 2.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 5, 7, 12, 18, 29, 42, 61, 85, 118, 164, 223, 299, 399, 530, 693, 888, 1157, 1488, 1901, 2403, 3044, 3807, 4783, 5935, 7368, 9097, 11197, 13721, 16806, 20441, 24868, 30133, 36494, 43895, 52880, 63424, 75900, 90609, 108088, 128404
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

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) = 2.

Crossrefs

Cf. A238616, A238658-A238660, A238662.

Programs

  • 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*)

Formula

a(n) + A238659(n) = A000041(n).

A238617 Number of partitions of n having standard deviation σ <= 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 10, 14, 15, 20, 23, 30, 33, 42, 44, 55, 58, 71, 78, 91, 99, 118, 124, 148, 155, 187, 202, 229, 244, 279, 290, 327, 344, 397, 427, 475, 501, 558, 597, 665, 714, 776, 824, 898, 948, 1032, 1084, 1245, 1308, 1395, 1452, 1606, 1692, 1807, 1919
Offset: 1

Views

Author

Clark Kimberling, Mar 01 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			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) = 9.

Crossrefs

Cf. A238616.

Programs

  • Maple
    b:= proc(n, i, m, s, c) `if`(n=0, `if`(s/c-(m/c)^2<=1, 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 = 55; g[n_] := g[n] = IntegerPartitions[n]; 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}]   (*A238616*)
Table[Count[g[n], p_ /; s[p] <= 1], {n, z}]  (*A238617*)
Table[Count[g[n], p_ /; s[p] == 1], {n, z}]  (*A238618*)
Table[Count[g[n], p_ /; s[p] > 1], {n, z}]   (*A238619*)
Table[Count[g[n], p_ /; s[p] >= 1], {n, z}]  (*A238620*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
ListPlot[Sort[t[30]]] (*plot of st.dev's of partitions of 30*)
b[n_, i_, m_, s_, c_] := b[n, i, m, s, c] = If[n == 0, If[s/c - (m/c)^2 <= 1, 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]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)

Formula

a(n) + A238619(n) = A000041(n).

A238618 Number of partitions of n having standard deviation σ = 1.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 4, 0, 4, 0, 7, 0, 1, 0, 16, 0, 1, 0, 10, 0, 11, 0, 10, 0, 1, 0, 26, 0, 1, 0, 26, 0, 24, 0, 13, 0, 1, 0, 60, 0, 5, 0, 17, 0, 19, 0, 52, 0, 1, 0, 117, 0, 1, 0, 36, 0, 46, 0, 23, 0, 29, 0, 160, 0, 1, 0, 30, 0, 61, 0, 140
Offset: 1

Views

Author

Clark Kimberling, Mar 01 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			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) = 1.

Crossrefs

Cf. A238616.

Programs

  • Maple
    b:= proc(n, i, m, s, c) `if`(n=0, `if`(s/c-(m/c)^2=1, 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 = 55; g[n_] := g[n] = IntegerPartitions[n]; 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}]   (*A238616*)
Table[Count[g[n], p_ /; s[p] <= 1], {n, z}]  (*A238617*)
Table[Count[g[n], p_ /; s[p] == 1], {n, z}]  (*A238618*)
Table[Count[g[n], p_ /; s[p] > 1], {n, z}]   (*A238619*)
Table[Count[g[n], p_ /; s[p] >= 1], {n, z}]  (*A238620*)
t[n_] := t[n] = N[Table[s[g[n][[k]]], {k, 1, PartitionsP[n]}]]
ListPlot[Sort[t[30]]] (*plot of st.dev's of partitions of 30*)
b[n_, i_, m_, s_, c_] := b[n, i, m, s, c] = If[n == 0, If[s/c - (m/c)^2 == 1, 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]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Nov 16 2015, after Alois P. Heinz *)

Extensions

a(56)-a(80) from Alois P. Heinz, Mar 11 2014

A238659 Number of partitions of n having standard deviation σ <= 2.

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 14, 20, 25, 35, 44, 59, 72, 93, 115, 146, 179, 221, 267, 328, 393, 472, 562, 687, 801, 948, 1109, 1315, 1521, 1797, 2059, 2414, 2775, 3213, 3686, 4256, 4831, 5574, 6317, 7205, 8089, 9279, 10381, 11751, 13234, 14949, 16666, 18869, 20986
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

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) = 20.

Crossrefs

Cf. A238616, A238658, A238660-A238662.

Programs

  • 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*)

Formula

a(n) + A238661(n) = A000041(n).

A239140 Number of strict partitions of n having standard deviation σ < 1.

Original entry on oeis.org

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

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Author

Clark Kimberling, Mar 11 2014

Keywords

Comments

Regarding standard deviation, see Comments at A238616.

Examples

			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.

Links

Crossrefs

Cf. A000009, A083039, A238616, A239141, A239142, A239143.
Column k=0 of A239228.

Programs

  • 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 *)
  • PARI
    A083039(n) = (1+!(n%2)+!(n%3));
A239140(n) = if(n<=3,1+(3==n),A083039(n-3)); \\ Antti Karttunen, May 24 2021

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

Extensions

A-number in the first formula corrected by Antti Karttunen, May 24 2021
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