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 11 results. Next

A237825 Number of partitions of n such that 3*(least part) = greatest part.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 3, 5, 5, 8, 9, 13, 14, 18, 20, 27, 28, 35, 38, 49, 51, 61, 66, 81, 86, 102, 109, 130, 136, 161, 172, 202, 214, 245, 264, 305, 323, 369, 395, 452, 480, 544, 580, 657, 703, 786, 842, 947, 1008, 1124, 1205, 1340, 1432, 1589, 1702, 1886, 2014
Offset: 1

Views

Author

Clark Kimberling, Feb 16 2014

Keywords

Examples

			a(7) = 3 counts these partitions:  331, 3211, 31111.
		

Crossrefs

Programs

  • Mathematica
    z = 64; q[n_] := q[n] = IntegerPartitions[n];
    Table[Count[q[n], p_ /; 3 Min[p] == Max[p]], {n, z}]     (* A237825*)
    Table[Count[q[n], p_ /; 4 Min[p] == Max[p]], {n, z}]     (* A237826 *)
    Table[Count[q[n], p_ /; 5 Min[p] == Max[p]], {n, z}]     (* A237827 *)
    Table[Count[q[n], p_ /; 2 Min[p] + 1 == Max[p]], {n, z}] (* A237828 *)
    Table[Count[q[n], p_ /; 2 Min[p] - 1 == Max[p]], {n, z}] (* A237829 *)
    Table[Count[IntegerPartitions[n],?(3#[[-1]]==#[[1]]&)],{n,60}] (* _Harvey P. Dale, May 14 2023 *)
    kmax = 57;
    Sum[x^(4 k)/Product[1 - x^j, {j, k, 3 k}], {k, 1, kmax}]/x + O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *)
  • PARI
    my(N=60, x='x+O('x^N)); concat([0, 0, 0], Vec(sum(k=1, N, x^(4*k)/prod(j=k, 3*k, 1-x^j)))) \\ Seiichi Manyama, May 14 2023

Formula

G.f.: Sum_{k>=1} x^(4*k)/Product_{j=k..3*k} (1-x^j). - Seiichi Manyama, May 14 2023
a(n) ~ c * A376815^sqrt(n) / sqrt(n), where c = 0.23036554... - Vaclav Kotesovec, Jun 14 2025

A237826 Number of partitions of n such that 4*(least part) = greatest part.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 9, 12, 16, 20, 26, 31, 38, 47, 55, 67, 78, 92, 106, 126, 145, 167, 190, 219, 247, 288, 320, 366, 410, 466, 520, 591, 654, 739, 820, 924, 1018, 1148, 1263, 1415, 1562, 1740, 1911, 2136, 2342, 2607, 2859, 3169, 3469, 3849, 4208
Offset: 1

Views

Author

Clark Kimberling, Feb 16 2014

Keywords

Examples

			a(8) = 3 counts these partitions:  431, 4211, 41111.
		

Crossrefs

Programs

  • Mathematica
    z = 64; q[n_] := q[n] = IntegerPartitions[n];
    Table[Count[q[n], p_ /; 3 Min[p] == Max[p]], {n, z}]     (* A237825*)
    Table[Count[q[n], p_ /; 4 Min[p] == Max[p]], {n, z}]     (* A237826 *)
    Table[Count[q[n], p_ /; 5 Min[p] == Max[p]], {n, z}]     (* A237827 *)
    Table[Count[q[n], p_ /; 2 Min[p] + 1 == Max[p]], {n, z}] (* A237828 *)
    Table[Count[q[n], p_ /; 2 Min[p] - 1 == Max[p]], {n, z}] (* A237829 *)
    Table[Count[IntegerPartitions[n],?(#[[1]]==4#[[-1]]&)],{n,60}] (* _Harvey P. Dale, Jun 15 2023 *)
    kmax = 55;
    Sum[x^(5k)/Product[1 - x^j, {j, k, 4 k}], {k, 1, kmax}]/x + O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *)
  • PARI
    my(N=60, x='x+O('x^N)); concat([0, 0, 0, 0], Vec(sum(k=1, N, x^(5*k)/prod(j=k, 4*k, 1-x^j)))) \\ Seiichi Manyama, May 14 2023

Formula

G.f.: Sum_{k>=1} x^(5*k)/Product_{j=k..4*k} (1-x^j). - Seiichi Manyama, May 14 2023
a(n) ~ c * d^sqrt(n) / sqrt(n), where d = 4.9219345... and c = 0.1699648... - Vaclav Kotesovec, Jun 19 2025

A237827 Number of partitions of n such that 5*(least part) = greatest part.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 13, 19, 24, 32, 39, 52, 61, 77, 93, 114, 133, 164, 188, 226, 261, 309, 353, 417, 471, 549, 622, 717, 808, 933, 1042, 1191, 1334, 1516, 1690, 1921, 2131, 2407, 2674, 3006, 3330, 3744, 4135, 4628, 5116, 5708, 6294, 7020
Offset: 1

Views

Author

Clark Kimberling, Feb 16 2014

Keywords

Examples

			a(8) = 2 counts these partitions:  521, 5111.
		

Crossrefs

Programs

  • Mathematica
    z = 64; q[n_] := q[n] = IntegerPartitions[n];
    Table[Count[q[n], p_ /; 3 Min[p] = = Max[p]], {n, z}]     (* A237825*)
    Table[Count[q[n], p_ /; 4 Min[p] = = Max[p]], {n, z}]     (* A237826 *)
    Table[Count[q[n], p_ /; 5 Min[p] = = Max[p]], {n, z}]     (* A237827 *)
    Table[Count[q[n], p_ /; 2 Min[p] + 1 = = Max[p]], {n, z}] (* A237828 *)
    Table[Count[q[n], p_ /; 2 Min[p] - 1 = = Max[p]], {n, z}] (* A237829 *)
    (* Second program: *)
    kmax = 54;
    Sum[x^(6 k)/Product[1 - x^j, {j, k, 5 k}], {k, 1, kmax}]/x + O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *)
  • PARI
    my(N=60, x='x+O('x^N)); concat([0, 0, 0, 0, 0], Vec(sum(k=1, N, x^(6*k)/prod(j=k, 5*k, 1-x^j)))) \\ Seiichi Manyama, May 14 2023

Formula

G.f.: Sum_{k>=1} x^(6*k)/Product_{j=k..5*k} (1-x^j). - Seiichi Manyama, May 14 2023
a(n) ~ c * d^sqrt(n) / sqrt(n), where d = 5.4930955... and c = 0.135509... - Vaclav Kotesovec, Jun 19 2025

A237829 Number of partitions of n such that 2*(least part) - 1 = greatest part.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 2, 2, 5, 3, 4, 5, 5, 6, 8, 6, 8, 10, 10, 10, 15, 12, 14, 17, 18, 20, 23, 21, 26, 29, 30, 31, 39, 38, 42, 46, 49, 52, 61, 60, 68, 74, 77, 83, 94, 95, 104, 112, 122, 128, 143, 144, 159, 172, 181, 192, 212, 219, 237, 253, 271, 285
Offset: 1

Views

Author

Clark Kimberling, Feb 16 2014

Keywords

Examples

			a(8) = 3 counts these partitions:  53, 332, 11111111.
		

Crossrefs

Programs

  • Mathematica
    z = 64; q[n_] := q[n] = IntegerPartitions[n];
    Table[Count[q[n], p_ /; 3 Min[p] == Max[p]], {n, z}]     (* A237825*)
    Table[Count[q[n], p_ /; 4 Min[p] == Max[p]], {n, z}]     (* A237826 *)
    Table[Count[q[n], p_ /; 5 Min[p] == Max[p]], {n, z}]     (* A237827 *)
    Table[Count[q[n], p_ /; 2 Min[p] + 1 == Max[p]], {n, z}] (* A237828 *)
    Table[Count[q[n], p_ /; 2 Min[p] - 1 == Max[p]], {n, z}] (* A237829 *)
    (* Second program: *)
    kmax = 64;
    Sum[x^(3k-1)/Product[1-x^j, {j, k, 2k-1}], {k, 1, kmax}]/x+1+O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *)
    nmax = 100; p = 1; s = x; Do[p = Simplify[p*(1 - x^(2*k - 1))*(1 - x^(2*k))/(1 - x^k)]; p = Normal[p + O[x]^(nmax + 1)]; s += x^(3*k - 1)/(1 - x^k)*(1 - x^(2*k))/p;, {k, 1, nmax}]; Rest[CoefficientList[Series[s, {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jun 18 2025 *)
  • PARI
    my(N=70, x='x+O('x^N)); Vec(x+sum(k=1, N, x^(3*k-1)/prod(j=k, 2*k-1, 1-x^j))) \\ Seiichi Manyama, May 17 2023

Formula

G.f.: x + Sum_{k>=1} x^(3*k-1)/Product_{j=k..2*k-1} (1-x^j). - Seiichi Manyama, May 17 2023
a(n) ~ exp(Pi*sqrt(2*n/15)) / (sqrt(2)* 5^(1/4) * phi^(3/2) * sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 20 2025

A239510 Number of partitions p of n such that if h = min(p), then h is an (h,0)-separator of p; see Comments.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 4, 5, 7, 11, 13, 18, 24, 30, 37, 48, 59, 73, 90, 109, 132, 163, 193, 233, 280, 334, 397, 475, 559, 663, 784, 924, 1085, 1279, 1494, 1751, 2049, 2392, 2784, 3248, 3769, 4382, 5081, 5887, 6808, 7879, 9087, 10486, 12083, 13910, 15988, 18384
Offset: 1

Views

Author

Clark Kimberling, Mar 24 2014

Keywords

Comments

Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)-separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)-separable if there is an ordering x, h, x, h, ..., x, h, where the number of h's on the ends is 1; next, p is (h,2)-separable if there is an ordering h, x, h, ..., x, h. Finally, p is h-separable if it is (h,i)-separable for i = 0,1,2.

Examples

			a(9) counts these 5 partitions: 612, 513, 414, 423, 312121.
		

Crossrefs

Programs

  • Mathematica
    z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239510 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239511 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p] - 1], {n, 1, z}]  (* A237828 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] - 1], {n, 1, z}]  (* A239513 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] - 1], {n, 1, z}] (* A239514 *)

A239511 Number of partitions p of n such that if h = 2*min(p), then h is an (h,0)-separator of p; see Comments.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 3, 4, 4, 5, 7, 9, 10, 11, 16, 17, 21, 26, 30, 38, 46, 53, 63, 76, 89, 106, 128, 149, 176, 210, 245, 287, 339, 392, 463, 542, 628, 733, 854, 989, 1150, 1336, 1542, 1782, 2063, 2373, 2736, 3155, 3620, 4162, 4783, 5476, 6275, 7185, 8210
Offset: 1

Views

Author

Clark Kimberling, Mar 24 2014

Keywords

Comments

Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)-separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)-separable if there is an ordering x, h, x, h, ..., x, h, where the number of h's on the ends is 1; next, p is (h,2)-separable if there is an ordering h, x, h, ..., x, h. Finally, p is h-separable if it is (h,i)-separable for i = 0,1,2.

Examples

			a(9) counts these 4 partitions: 612, 513, 324, 31212.
		

Crossrefs

Programs

  • Mathematica
    z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239510 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239511 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p] - 1], {n, 1, z}]  (* A237828 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] - 1], {n, 1, z}]  (* A239513 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] - 1], {n, 1, z}] (* A239514 *)

A239513 Number of partitions p of n such that if h = (number of parts of p), then h is an (h,0)-separator of p; see Comments.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 5, 5, 7, 8, 9, 10, 12, 13, 15, 17, 19, 21, 25, 27, 31, 35, 40, 44, 50, 55, 62, 68, 76, 83, 93, 101, 112, 122, 136, 147, 163, 177, 196, 213, 235, 255, 281, 305, 335, 363, 398, 431, 471, 510, 556, 601, 654, 706, 768, 828
Offset: 1

Views

Author

Clark Kimberling, Mar 24 2014

Keywords

Comments

Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)-separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)-separable if there is an ordering x, h, x, h, ..., x, h, where the number of h's on the ends is 1; next, p is (h,2)-separable if there is an ordering h, x, h, ..., x, h. Finally, p is h-separable if it is (h,i)-separable for i = 0,1,2.

Examples

			a(13) counts these 5 partitions: 931, 832, 634, 535, 15151.
		

Crossrefs

Programs

  • Mathematica
    z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239510 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239511 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p] - 1], {n, 1, z}]  (* A237828 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] - 1], {n, 1, z}]  (* A239513 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] - 1], {n, 1, z}] (* A239514 *)

A239514 Number of partitions p of n such that if h = max(p) - min(p), then h is an (h,0)-separator of p; see Comments.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 2, 2, 3, 2, 4, 2, 7, 6, 7, 6, 10, 7, 14, 12, 18, 12, 22, 18, 23, 23, 31, 29, 42, 33, 45, 42, 54, 49, 68, 62, 78, 76, 95, 87, 110, 102, 124, 128, 150, 141, 178, 174, 203, 203, 237, 228, 272, 269, 308, 318, 360, 356, 422, 420, 472, 482
Offset: 1

Views

Author

Clark Kimberling, Mar 24 2014

Keywords

Comments

Suppose that p is a partition of n into 2 or more parts and that h is a part of p. Then p is (h,0)-separable if there is an ordering x, h, x, h, ..., h, x of the parts of p, where each x represents any part of p except h. Here, the number of h's on the ends of the ordering is 0. Similarly, p is (h,1)-separable if there is an ordering x, h, x, h, ..., x, h, where the number of h's on the ends is 1; next, p is (h,2)-separable if there is an ordering h, x, h, ..., x, h. Finally, p is h-separable if it is (h,i)-separable for i = 0,1,2.

Crossrefs

Programs

  • Mathematica
    z = 75; Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239510 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, 2 Min[p]] == Length[p] - 1], {n, 1, z}]  (* A239511 *)
    Table[Count[IntegerPartitions[n], p_ /; 2 Count[p, Max[p]] == Length[p] - 1], {n, 1, z}]  (* A237828 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Length[p]] == Length[p] - 1], {n, 1, z}]  (* A239513 *)
    Table[Count[Rest[IntegerPartitions[n]], p_ /; 2 Count[p, Max[p] - Min[p]] == Length[p] - 1], {n, 1, z}] (* A239514 *)

Formula

a(12) counts these partitions: 615, 642, 43131, 3121212.

A363075 Number of partitions of n such that 3*(least part) + 1 = greatest part.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 3, 6, 6, 10, 12, 18, 20, 27, 32, 42, 47, 59, 67, 85, 94, 113, 126, 152, 169, 198, 220, 257, 282, 326, 359, 413, 452, 512, 563, 639, 695, 781, 853, 958, 1041, 1161, 1261, 1402, 1524, 1685, 1827, 2021, 2186, 2407, 2604, 2861, 3088, 3385, 3657, 4002, 4316, 4704, 5069, 5531
Offset: 1

Views

Author

Seiichi Manyama, May 17 2023

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 100; p = 1; s = 0; Do[p = Simplify[p*(1 - x^(3*k - 2))*(1 - x^(3*k - 1))*(1 - x^(3*k))/(1 - x^k)]; p = Normal[p + O[x]^(nmax + 1)]; s += x^(4*k + 1)/(1 - x^k)/(1 - x^(3*k + 1))/p;, {k, 1, nmax}]; Rest[CoefficientList[Series[s, {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jun 19 2025 *)
  • PARI
    my(N=70, x='x+O('x^N)); concat([0, 0, 0, 0], Vec(sum(k=1, N, x^(4*k+1)/prod(j=k, 3*k+1, 1-x^j))))

Formula

G.f.: Sum_{k>=1} x^(4*k+1)/Product_{j=k..3*k+1} (1-x^j).
a(n) ~ c * A376815^sqrt(n) / sqrt(n), where c = 0.33761... - Vaclav Kotesovec, Jun 20 2025

A363076 Number of partitions of n such that 4*(least part) + 1 = greatest part.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 8, 10, 14, 19, 25, 33, 41, 51, 65, 79, 97, 116, 140, 165, 198, 233, 272, 316, 369, 422, 493, 561, 643, 731, 835, 943, 1072, 1205, 1359, 1524, 1717, 1911, 2147, 2387, 2665, 2960, 3295, 3640, 4049, 4469, 4950, 5455, 6028, 6622, 7310, 8024, 8826, 9676, 10632, 11627, 12765
Offset: 1

Views

Author

Seiichi Manyama, May 17 2023

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 100; p = 1; s = 0; Do[p = Simplify[p*(1 - x^(4*k - 3))*(1 - x^(4*k - 2))*(1 - x^(4*k - 1))*(1 - x^(4*k))/(1 - x^k)]; p = Normal[p + O[x]^(nmax + 1)]; s += x^(5*k + 1)/(1 - x^k)/(1 - x^(4*k + 1))/p;, {k, 1, nmax}]; Rest[CoefficientList[Series[s, {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jun 19 2025 *)
  • PARI
    my(N=70, x='x+O('x^N)); concat([0, 0, 0, 0, 0], Vec(sum(k=1, N, x^(5*k+1)/prod(j=k, 4*k+1, 1-x^j))))

Formula

G.f.: Sum_{k>=1} x^(5*k+1)/Product_{j=k..4*k+1} (1-x^j).
Showing 1-10 of 11 results. Next