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 1-4 of 4 results.

A237828 Number of partitions of n such that 2*(least part) + 1 = greatest part.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 4, 4, 6, 9, 10, 12, 17, 18, 22, 27, 31, 34, 42, 45, 53, 61, 66, 72, 86, 92, 103, 113, 125, 135, 154, 163, 180, 197, 213, 229, 257, 271, 294, 318, 346, 368, 404, 426, 463, 497, 532, 564, 616, 651, 700, 747, 798, 844, 912, 962, 1033, 1097, 1167, 1231, 1327, 1397, 1486, 1576, 1677
Offset: 1

Views

Author

Clark Kimberling, Feb 16 2014

Keywords

Comments

Also, the number of partitions p of n such that if h = max(p), then h is an (h,0)-separator of p; for example, a(10) counts these 9 partitions: 181, 271, 361, 262, 451, 352, 343, 23131, 1212121. - Clark Kimberling, Mar 24 2014

Examples

			a(8) = 4 counts these partitions:  3311, 3221, 32111, 311111.

Links

Crossrefs

Cf. A118096, A237829.
Cf. A049820, A237757, A237825, A237826, A237827, A000041.

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],?(2*Min[#]+1==Max[#]&)],{n,60}] (* _Harvey P. Dale, Jun 25 2017 *)
kmax = 65;
Sum[x^(3k+1)/Product[1-x^j, {j, k, 2k+1}], {k, 1, kmax}]/x + O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *)
nmax = 100; p = 1; s = 0; 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 + 1))/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)); concat([0, 0, 0], Vec(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.: Sum_{k>=1} x^(3*k+1)/Product_{j=k..2*k+1} (1-x^j). - Seiichi Manyama, May 17 2023
a(n) ~ sqrt(phi) * exp(Pi*sqrt(2*n/15)) / (sqrt(2)* 5^(1/4) * sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 20 2025

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.

Links

Crossrefs

Cf. A000041, A117086, A237757, A237828, A237829.
Cf. A118096, A237826, A237827.

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.

Links

Crossrefs

Cf. A000041, A117086, A237757, A237828, A237829.
Cf. A118096, A237825, A237827.

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.

Links

Crossrefs

Cf. A000041, A117086, A237757, A237828, A237829.
Cf. A118096, A237825, A237826.

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
Showing 1-4 of 4 results.

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