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.

Previous Showing 11-17 of 17 results.

A238655 Number of partitions of n having standard deviation σ > 3.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 5, 10, 15, 23, 33, 49, 68, 99, 133, 179, 246, 334, 432, 583, 756, 974, 1261, 1618, 2076, 2657, 3336, 4228, 5270, 6592, 8190, 10182, 12567, 15533, 19008, 23307, 28410, 34622, 42041, 50959, 61487, 74259, 88734, 106666, 127587
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			There are 30 partitions of 9, whose standard deviations are given by these approximations:  0., 3.5, 2.5, 2.82843, 1.5, 2.16025, 2.16506, 0.5, 1.63299, 1.41421, 1.63936, 1.6, 1.41421, 0.816497, 1.29904, 1.08972, 1.16619, 1.11803, 0., 0.829156, 0.979796, 0.433013, 0.748331, 0.763763, 0.699854, 0.4, 0.5, 0.451754, 0.330719, 0, so that a(9) = 1.
		

Crossrefs

Programs

  • Mathematica
    z = 53; 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] > 3], {n, z}]   (*A238655*)
    Table[Count[g[n], p_ /; s[p] > 4], {n, z}]   (*A238656*)
    Table[Count[g[n], p_ /; s[p] > 5], {n, z}]   (*A238657*)
    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*)

A239141 Number of strict partitions of n having standard deviation <= 1.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2
Offset: 1

Views

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 0.0, 3.5, 2.5, 1.5, 2.16025, 0.5, 1.63299, 0.816497, so that a(9) = 3.
		

Crossrefs

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[{0, 0, 1},{2, 2, 3},83]] (* Ray Chandler, Aug 25 2015 *)
  • PARI
    A239141(n) = (1+(n>3)+!(n%3)); \\ Antti Karttunen, May 24 2021

Formula

a(n) + A239142(n) = A000009(n) for n >= 1.
G.f.: -(x^5 + x^4 + x^3 + 2*x^2 + x + 1)*x / ((x-1)*(x^2 + x + 1)). - Alois P. Heinz, Mar 14 2014

A239142 Number of strict partitions of n having standard deviation sigma > 1.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 3, 4, 5, 8, 10, 12, 16, 20, 24, 30, 36, 43, 52, 62, 73, 87, 102, 119, 140, 163, 189, 220, 254, 293, 338, 388, 445, 510, 583, 665, 758, 862, 979, 1111, 1258, 1423, 1608, 1814, 2045, 2302, 2588, 2907, 3262, 3656, 4094, 4580, 5118, 5715, 6376
Offset: 1

Views

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) = 5.
		

Crossrefs

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 *)

Formula

a(n) + A239141(n) = A000009(n) for n >=1.
G.f.: Product_{m>=1} (1+x^m) -1 +(x^5+x^4+x^3+2*x^2+x+1)*x / ((x-1)*(x^2+x+1)). - Alois P. Heinz, Mar 14 2014

A239143 Number of strict partitions of n having standard deviation σ >= 1.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 5, 5, 9, 10, 13, 16, 21, 24, 31, 36, 44, 52, 63, 73, 88, 102, 120, 140, 164, 189, 221, 254, 294, 338, 389, 445, 511, 583, 666, 758, 863, 979, 1112, 1258, 1424, 1608, 1815, 2045, 2303, 2588, 2908, 3262, 3657, 4094, 4581, 5118, 5716, 6376
Offset: 1

Views

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) = 5.
		

Crossrefs

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 *)

Formula

a(n) + A239140(n) = A000009(n) for n >=1.
G.f.: Product_{m>=1} (1+x^m) -1 +(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

A238656 Number of partitions of n having standard deviation σ > 4.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 4, 5, 9, 14, 19, 28, 41, 57, 80, 109, 149, 199, 265, 351, 457, 599, 780, 1011, 1299, 1664, 2121, 2682, 3377, 4252, 5345, 6660, 8279, 10277, 12733, 15596, 19245, 23556, 28761, 35066, 42723, 51615, 62657, 75494, 90978
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			There are 30 partitions of 9, whose standard deviations are given by these approximations:  0., 3.5, 2.5, 2.82843, 1.5, 2.16025, 2.16506, 0.5, 1.63299, 1.41421, 1.63936, 1.6, 1.41421, 0.816497, 1.29904, 1.08972, 1.16619, 1.11803, 0., 0.829156, 0.979796, 0.433013, 0.748331, 0.763763, 0.699854, 0.4, 0.5, 0.451754, 0.330719, 0, so that a(9) = 0.
		

Crossrefs

Programs

  • Mathematica
    z = 53; 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] > 3], {n, z}]   (*A238655*)
    Table[Count[g[n], p_ /; s[p] > 4], {n, z}]   (*A238656*)
    Table[Count[g[n], p_ /; s[p] > 5], {n, z}]   (*A238657*)
    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*)

A238657 Number of partitions of n having standard deviation σ > 5.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 5, 9, 11, 16, 25, 34, 45, 64, 87, 121, 160, 212, 279, 369, 481, 614, 797, 1027, 1308, 1670, 2102, 2661, 3345, 4189, 5224, 6494, 8069, 9982, 12281, 15093, 18508, 22731, 27564, 33639, 40757, 49496, 59838, 72228
Offset: 1

Views

Author

Clark Kimberling, Mar 03 2014

Keywords

Comments

Regarding "standard deviation" see Comments at A238616.

Examples

			There are 30 partitions of 9, whose standard deviations are given by these approximations:  0., 3.5, 2.5, 2.82843, 1.5, 2.16025, 2.16506, 0.5, 1.63299, 1.41421, 1.63936, 1.6, 1.41421, 0.816497, 1.29904, 1.08972, 1.16619, 1.11803, 0., 0.829156, 0.979796, 0.433013, 0.748331, 0.763763, 0.699854, 0.4, 0.5, 0.451754, 0.330719, 0, so that a(9) = 0.
		

Crossrefs

Programs

  • Mathematica
    z = 53; 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] > 3], {n, z}]   (*A238655*)
    Table[Count[g[n], p_ /; s[p] > 4], {n, z}]   (*A238656*)
    Table[Count[g[n], p_ /; s[p] > 5], {n, z}]   (*A238657*)
    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*)

A239223 Number T(n,k) of partitions of n with standard deviation σ in the half-open interval [k,k+1); triangle T(n,k), n>=1, 0<=k<=max(0,floor(n/2)-1), read by rows.

Original entry on oeis.org

1, 2, 3, 4, 1, 6, 1, 8, 2, 1, 10, 4, 1, 12, 7, 2, 1, 15, 10, 4, 1, 19, 14, 6, 2, 1, 23, 21, 7, 4, 1, 25, 32, 14, 3, 2, 1, 33, 39, 19, 6, 3, 1, 41, 51, 27, 10, 3, 2, 1, 44, 70, 39, 13, 7, 2, 1, 51, 92, 52, 21, 9, 3, 2, 1, 58, 121, 69, 30, 10, 6, 2, 1, 67, 149
Offset: 1

Views

Author

Alois P. Heinz, Mar 12 2014

Keywords

Examples

			Triangle T(n,k) begins:
   1;
   2;
   3;
   4,  1;
   6,  1;
   8,  2,  1;
  10,  4,  1;
  12,  7,  2, 1;
  15, 10,  4, 1;
  19, 14,  6, 2, 1;
  23, 21,  7, 4, 1;
  25, 32, 14, 3, 2, 1;
		

Crossrefs

Column k=0 gives A238616.
Row sums give A000041.
Maximal index in row n is A140106(n).
Cf. A239228.

Programs

  • Maple
    b:= proc(n, i, m, s, c) `if`(n=0, x^floor(sqrt(s/c-(m/c)^2)),
          `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:
    T:= n->(p->seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, 0$3)):
    seq(T(n), n=1..18);
  • Mathematica
    b[n_, i_, m_, s_, c_] := b[n, i, m, s, c] = If[n==0, x^Floor[Sqrt[s/c - (m/c)^2]], 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}]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n, 0, 0, 0]]; Table[T[n], {n, 1, 18}] // Flatten (* Jean-François Alcover, Nov 17 2015, translated from Maple *)
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