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 24 results. Next

A053263 Coefficients of the '5th-order' mock theta function chi_1(q).

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 6, 5, 7, 8, 9, 9, 12, 12, 15, 15, 18, 19, 23, 23, 27, 30, 33, 34, 41, 42, 49, 51, 57, 61, 69, 72, 81, 87, 96, 100, 113, 119, 132, 140, 153, 163, 180, 188, 208, 221, 240, 253, 278, 294, 319, 339, 366, 388, 422, 443, 481, 510, 549, 580, 626, 662
Offset: 0

Views

Author

Dean Hickerson, Dec 19 1999

Keywords

Comments

The rank of a partition is its largest part minus the number of parts.
Number of partitions of n such that 2*(least part) > greatest part. - Clark Kimberling, Feb 16 2014
Also the number of partitions of n with the same median as maximum. These are conjugate to the partitions described above. For minimum instead of maximum we have A361860. - Gus Wiseman, Apr 23 2023

Examples

			From _Gus Wiseman_, Apr 20 2023: (Start)
The a(1) = 1 through a(8) = 6 partitions such that 2*(minimum) > (maximum):
  (1)  (2)   (3)    (4)     (5)      (6)       (7)        (8)
       (11)  (111)  (22)    (32)     (33)      (43)       (44)
                    (1111)  (11111)  (222)     (322)      (53)
                                     (111111)  (1111111)  (332)
                                                          (2222)
                                                          (11111111)
The a(1) = 1 through a(8) = 6 partitions such that (median) = (maximum):
  (1)  (2)   (3)    (4)     (5)      (6)       (7)        (8)
       (11)  (111)  (22)    (221)    (33)      (331)      (44)
                    (1111)  (11111)  (222)     (2221)     (332)
                                     (111111)  (1111111)  (2222)
                                                          (22211)
                                                          (11111111)
(End)
		

References

  • Srinivasa Ramanujan, Collected Papers, Chelsea, New York, 1962, pp. 354-355
  • Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, pp. 20, 25

Crossrefs

Other '5th-order' mock theta functions are at A053256, A053257, A053258, A053259, A053260, A053261, A053262, A053264, A053265, A053266, A053267.
A000041 counts integer partitions, strict A000009, odd-length A027193.
A359893 and A359901 count partitions by median.

Programs

  • Mathematica
    1+Series[Sum[q^(2n+1)(1+q^n)/Product[1-q^k, {k, n+1, 2n+1}], {n, 0, 49}], {q, 0, 100}]
    (* Also: *)
    Table[Count[ IntegerPartitions[n], p_ /; 2 Min[p] > Max[p]], {n, 40}]
    (* Clark Kimberling, Feb 16 2014 *)
    nmax = 100; CoefficientList[Series[1 + Sum[x^(2*k+1)*(1+x^k) / Product[1-x^j, {j, k+1, 2*k+1}], {k, 0, Floor[nmax/2]}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 12 2019 *)

Formula

G.f.: chi_1(q) = Sum_{n>=0} q^n/((1-q^(n+1))(1-q^(n+2))...(1-q^(2n+1))).
G.f.: chi_1(q) = 1 + Sum_{n>=0} q^(2n+1) (1+q^n)/((1-q^(n+1))(1-q^(n+2))...(1-q^(2n+1))).
a(n) is twice the number of partitions of 5n+3 with rank == 2 (mod 5) minus number with rank == 0 or 1 (mod 5).
a(n) - 1 is the number of partitions of n with unique smallest part and all other parts <= one plus twice the smallest part.
a(n) ~ sqrt(phi/2) * exp(Pi*sqrt(2*n/15)) / (5^(1/4)*sqrt(n)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jun 16 2019

A237753 Number of partitions of n such that 2*(greatest part) = (number of parts).

Original entry on oeis.org

0, 1, 0, 0, 1, 1, 1, 2, 1, 2, 3, 4, 5, 7, 7, 9, 12, 15, 17, 23, 27, 34, 42, 50, 60, 75, 87, 106, 128, 154, 182, 222, 260, 311, 369, 437, 515, 613, 716, 845, 993, 1166, 1361, 1599, 1861, 2176, 2534, 2950, 3422, 3983, 4605, 5339, 6174, 7136, 8227, 9500, 10928
Offset: 1

Views

Author

Clark Kimberling, Feb 13 2014

Keywords

Comments

Also, the number of partitions of n such that (greatest part) = 2*(number of parts); hence, the number of partitions of n such that (rank + greatest part) = 0.

Examples

			a(8) = 2 counts these partitions:  311111, 2222.
		

Crossrefs

Programs

  • Mathematica
    z = 50; Table[Count[IntegerPartitions[n], p_ /; 2 Max[p] = = Length[p]], {n, z}]
    (* or *)
    nmax = 100; Rest[CoefficientList[Series[Sum[x^(3*k-1) * Product[(1 - x^(2*k+j-1)) / (1 - x^j), {j, 1, k-1}], {k, 1, nmax/3 + 1}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 15 2024 *)
    nmax = 100; p = x; s = x; Do[p = Normal[Series[p*x^3*(1 - x^(3*k - 1))*(1 - x^(3*k))*(1 - x^(3*k + 1))/((1 - x^(2*k + 1))*(1 - x^(2*k))*(1 - x^k)), {x, 0, nmax}]]; s += p;, {k, 1, nmax/3 + 1}]; Take[CoefficientList[s, x], nmax] (* Vaclav Kotesovec, Oct 16 2024 *)
  • PARI
    my(N=66, x='x+O('x^N)); concat(0, Vec(sum(k=1, N, x^(3*k-1)*prod(j=1, k-1, (1-x^(2*k+j-1))/(1-x^j))))) \\ Seiichi Manyama, Jan 24 2022

Formula

G.f.: Sum_{k>=1} x^(3*k-1) * Product_{j=1..k-1} (1-x^(2*k+j-1))/(1-x^j). - Seiichi Manyama, Jan 24 2022
a(n) ~ Pi^2 * exp(Pi*sqrt(2*n/3)) / (4 * 3^(3/2) * n^2). - Vaclav Kotesovec, Oct 17 2024

A237755 Number of partitions of n such that 2*(greatest part) >= (number of parts).

Original entry on oeis.org

1, 2, 2, 4, 6, 9, 12, 18, 24, 34, 46, 63, 83, 111, 144, 190, 245, 318, 405, 520, 657, 833, 1045, 1312, 1634, 2036, 2517, 3114, 3829, 4705, 5751, 7027, 8544, 10381, 12564, 15190, 18301, 22026, 26425, 31669, 37849, 45180, 53796, 63983, 75923, 89987, 106435
Offset: 1

Views

Author

Clark Kimberling, Feb 13 2014

Keywords

Comments

Also, the number of partitions of n such that (greatest part) <= 2*(number of parts); hence, the number of partitions of n such that (rank + greatest part) >= 0.

Examples

			a(6) = 9 counts all of the 11 partitions of 6 except these:  21111, 111111.
		

Crossrefs

Programs

  • Mathematica
    z = 50; Table[Count[IntegerPartitions[n], p_ /; 2 Max[p] >= Length[p]], {n, z}]
  • PARI
    {a(n) = my(A); A = sum(m=0,n,x^m*prod(k=1,m,(1-x^(2*m+k-1))/(1-x^k +x*O(x^n)))); polcoeff(A,n)}
    for(n=1,60,print1(a(n),", ")) \\ Paul D. Hanna, Aug 03 2015

Formula

a(n) = A000041(n) - A237751(n).
G.f.: Sum_{n>=1} x^n * Product_{k=1..n} (1 - x^(2*n+k-1))/(1 - x^k). - Paul D. Hanna, Aug 03 2015

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.
		

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],?(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.
		

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

A237751 Number of partitions of n such that 2*(greatest part) < (number of parts).

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 3, 4, 6, 8, 10, 14, 18, 24, 32, 41, 52, 67, 85, 107, 135, 169, 210, 263, 324, 400, 493, 604, 736, 899, 1091, 1322, 1599, 1929, 2319, 2787, 3336, 3989, 4760, 5669, 6734, 7994, 9465, 11192, 13211, 15571, 18319, 21531, 25257, 29594, 34626
Offset: 1

Views

Author

Clark Kimberling, Feb 13 2014

Keywords

Comments

Also, the number of partitions of n such that (greatest part) > 2*(number of parts); hence, the number of partitions of n such that (rank + greatest part) < 0.

Examples

			a(6) = 2 counts these partitions:  21111, 111111.
		

Crossrefs

Programs

  • Mathematica
    z = 55; Table[Count[IntegerPartitions[n], p_ /; 2 Max[p] < Length[p]], {n, z}]

Formula

a(n) = A000041(n) - A237755(n).

A237752 Number of partitions of n such that 2*(greatest part) <= (number of parts).

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 4, 6, 7, 10, 13, 18, 23, 31, 39, 50, 64, 82, 102, 130, 162, 203, 252, 313, 384, 475, 580, 710, 864, 1053, 1273, 1544, 1859, 2240, 2688, 3224, 3851, 4602, 5476, 6514, 7727, 9160, 10826, 12791, 15072, 17747, 20853, 24481, 28679, 33577, 39231
Offset: 1

Views

Author

Clark Kimberling, Feb 13 2014

Keywords

Comments

Also, the number of partitions of n such that (greatest part) >= 2*(number of parts); hence, the number of partitions of n such that (rank + greatest part) <= 0.
Also, the number of partitions p of n such that max(max(p), 2*(number of parts of p)) is a part of p.

Examples

			The partitions of 6 that do not qualify are 22311, 21111, 111111, so that a(6) = 11 - 3 = 8.
		

Crossrefs

Programs

  • Mathematica
    z = 50; Table[Count[IntegerPartitions[n], p_ /; 2 Max[p] <= Length[p]], {n, z}]
    (* also *)
    Table[Count[IntegerPartitions[n], p_ /; MemberQ[p, Max[Max[p],2*Length[p]]]], {n, 50}]

Formula

a(n) = A000041(n) - A237754(n).

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

A350889 Triangle T(n,k), n >= 1, 1 <= k <= n, read by rows, where T(n,k) is the number of partitions of n such that k*(smallest part) = (number of parts).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 3, 3, 2, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 2, 4, 5, 5, 3, 2, 1, 1, 2, 3, 4, 7, 6, 5, 3, 2, 1, 1, 3, 4, 5, 8, 9, 7, 5, 3, 2, 1, 1, 3, 5, 6, 10, 11, 10, 7, 5, 3, 2, 1, 1, 4, 6, 7, 12, 15, 13, 11, 7, 5, 3, 2, 1, 1, 4, 8, 8, 14, 18, 18, 14, 11, 7, 5, 3, 2, 1, 1
Offset: 1

Views

Author

Seiichi Manyama, Jan 21 2022

Keywords

Comments

Column k is asymptotic to r^2 * (k*log(r)^2 + polylog(2, r^2))^(1/4) * exp(2*sqrt((k*log(r)^2 + polylog(2, r^2))*n)) / (2*sqrt(Pi*k*(k - (k-2)*r^2)) * n^(3/4)), where r is the positive real root of the equation r^2 = 1 - r^k. - Vaclav Kotesovec, Oct 14 2024

Examples

			Triangle begins:
  1;
  0, 1;
  0, 1, 1;
  1, 1, 1, 1;
  1, 1, 2, 1, 1;
  1, 1, 2, 2, 1, 1;
  1, 1, 3, 3, 2, 1, 1;
  1, 2, 3, 4, 3, 2, 1, 1;
  2, 2, 4, 5, 5, 3, 2, 1, 1;
  2, 3, 4, 7, 6, 5, 3, 2, 1, 1;
  3, 4, 5, 8, 9, 7, 5, 3, 2, 1, 1;
		

Crossrefs

Row sums give A168657.

Programs

  • PARI
    T(n, k) = polcoef(sum(i=1, sqrtint(n\k), x^(k*i^2)/prod(j=1, k*i-1, 1-x^j+x*O(x^n))), n);
    
  • Ruby
    def partition(n, min, max)
      return [[]] if n == 0
      [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i).map{|rest| [i, *rest]}}
    end
    def A(n)
      a = Array.new(n, 0)
      partition(n, 1, n).each{|ary|
        (1..n).each{|i|
          a[i - 1] += 1 if i * ary[-1] == ary.size
        }
      }
      a
    end
    def A350889(n)
      (1..n).map{|i| A(i)}.flatten
    end
    p A350889(14)

Formula

G.f. of column k: Sum_{i>=1} x^(k*i^2)/Product_{j=1..k*i-1} (1-x^j).
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