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 11-20 of 21 results. Next

A206558 Number of 8's in the last section of the set of partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 8, 8, 13, 15, 23, 26, 38, 45, 63, 74, 101, 120, 160, 191, 248, 298, 383, 457, 579, 694, 868, 1038, 1287, 1536, 1890, 2251, 2746, 3267, 3962, 4698, 5665, 6706, 8043, 9496, 11337, 13354, 15876, 18657, 22089
Offset: 1

Views

Author

Omar E. Pol, Feb 09 2012

Keywords

Comments

Zero together with the first differences of A024792. Also number of occurrences of 8 in all partitions of n that do not contain 1 as a part. For the definition of "last section of n" see A135010. It appears that the sums of eight successive terms give the partition numbers A000041.

Crossrefs

Programs

  • Sage
    A206558 = lambda n: sum(list(p).count(8) for p in Partitions(n) if 1 not in p)

Formula

It appears that A000041(n) = Sum_{j=1..8} a(n+j), n >= 0.

A206559 Number of 9's in the last section of the set of partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 9, 12, 15, 22, 26, 36, 45, 59, 73, 97, 117, 152, 187, 236, 289, 365, 442, 551, 671, 825, 999, 1226, 1474, 1796, 2159, 2609, 3124, 3765, 4485, 5377, 6396, 7627, 9041, 10750, 12696, 15038, 17724, 20909
Offset: 1

Views

Author

Omar E. Pol, Feb 09 2012

Keywords

Comments

Zero together with the first differences of A024793. Also number of occurrences of 9 in all partitions of n that do not contain 1 as a part. For the definition of "last section of n" see A135010. It appears that the sums of nine successive terms give the partition numbers A000041.

Crossrefs

Programs

  • Sage
    A206559 = lambda n: sum(list(p).count(9) for p in Partitions(n) if 1 not in p)

Formula

It appears that A000041(n) = Sum_{j=1..9} a(n+j), n >= 0.

A240058 Number of partitions of n such that m(1) = m(3), where m = multiplicity.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 4, 2, 8, 5, 12, 9, 21, 17, 32, 29, 52, 49, 79, 79, 123, 126, 184, 195, 278, 299, 409, 449, 603, 668, 874, 979, 1263, 1423, 1803, 2045, 2563, 2916, 3608, 4121, 5056, 5783, 7029, 8055, 9725, 11151, 13366, 15337, 18285, 20979, 24871, 28535
Offset: 1

Views

Author

Clark Kimberling, Mar 31 2014

Keywords

Examples

			a(6) counts these 4 partitions:  6, 42, 321, 222.
		

Crossrefs

Programs

  • Mathematica
    z = 60; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Count[p, 1] < Count[p, 3]], {n, 0, z}]  (* A182714 *)
    t2 = Table[Count[f[n], p_ /; Count[p, 1] <= Count[p, 3]], {n, 0, z}] (* A182714(n+3) *)
    t3 = Table[Count[f[n], p_ /; Count[p, 1] == Count[p, 3]], {n, 0, z}] (* A240058 *)
    t4 = Table[Count[f[n], p_ /; Count[p, 1] > Count[p, 3]], {n, 0, z}]  (* A240059 *)
    t5 = Table[Count[f[n], p_ /; Count[p, 1] >= Count[p, 3]], {n, 0, z}] (* A240059(n+1) *)

Formula

a(n) = A182714(n+3) - A182714(n) = A240059(n+1) - A240059(n) for n >= 0.

A240063 Number of partitions of n such that m(2) < m(3), where m = multiplicity.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 3, 5, 7, 9, 13, 18, 25, 33, 44, 58, 76, 100, 129, 165, 212, 269, 342, 431, 540, 675, 842, 1045, 1292, 1592, 1957, 2397, 2931, 3569, 4337, 5258, 6358, 7671, 9236, 11091, 13296, 15906, 18994, 22634, 26927, 31974, 37907, 44867, 53017, 62547
Offset: 0

Views

Author

Clark Kimberling, Mar 31 2014

Keywords

Examples

			a(8) counts these 5 partitions:  53, 431, 332, 3311, 311111.
		

Crossrefs

Programs

  • Mathematica
    z = 60; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Count[p, 2] < Count[p, 3]], {n, 0, z}]  (* A240063 *)
    t2 = Table[Count[f[n], p_ /; Count[p, 2] <= Count[p, 3]], {n, 0, z}] (* A240063(n+3) *)
    t3 = Table[Count[f[n], p_ /; Count[p, 2] == Count[p, 3]], {n, 0, z}] (* A240064 *)
    t4 = Table[Count[f[n], p_ /; Count[p, 2] > Count[p, 3]], {n, 0, z}]  (* A240065 *)
    t5 = Table[Count[f[n], p_ /; Count[p, 2] >= Count[p, 3]], {n, 0, z}] (* A240065(n+2) *)
    nmax = 50; CoefficientList[Series[x^3*(1 + x)/((1 + x + x^2 + x^3 + x^4) * Product[1-x^k, {k, 1, nmax}]), {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 06 2021 *)

Formula

a(n) + A240064(n) + A240065(n) = A000041(n) for n >= 0.
From Vaclav Kotesovec, Oct 06 2021: (Start)
G.f.: x^3*(1 + x)/((1 + x + x^2 + x^3 + x^4) * Product_{k>=1} (1 - x^k)).
a(n) ~ exp(Pi*sqrt(2*n/3)) / (10*n*sqrt(3)).
a(n) ~ 2*A000041(n)/5. (End)

A240064 Number of partitions of n such that m(2) = m(3), where m = multiplicity.

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 5, 6, 8, 11, 16, 20, 26, 33, 43, 56, 71, 89, 112, 140, 177, 219, 271, 333, 411, 505, 617, 750, 912, 1105, 1339, 1612, 1940, 2327, 2789, 3334, 3978, 4733, 5625, 6670, 7903, 9338, 11021, 12980, 15273, 17940, 21043, 24640, 28822, 33661, 39273
Offset: 0

Views

Author

Clark Kimberling, Mar 31 2014

Keywords

Examples

			a(6) counts these 5 partitions:  6, 51, 411, 321, 111111.
		

Crossrefs

Programs

  • Mathematica
    z = 60; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Count[p, 2] < Count[p, 3]], {n, 0, z}]  (* A240063 *)
    t2 = Table[Count[f[n], p_ /; Count[p, 2] <= Count[p, 3]], {n, 0, z}] (* A240063(n+3) *)
    t3 = Table[Count[f[n], p_ /; Count[p, 2] == Count[p, 3]], {n, 0, z}] (* A240064 *)
    t4 = Table[Count[f[n], p_ /; Count[p, 2] > Count[p, 3]], {n, 0, z}]  (* A240065 *)
    t5 = Table[Count[f[n], p_ /; Count[p, 2] >= Count[p, 3]], {n, 0, z}] (* A240065(n+2) *)
  • PARI
    seq(n) = Vec((1-x^2)*(1-x^3)/((1-x^5)*eta(x + O(x*x^n)))) \\ Andrew Howroyd, Jan 01 2025

Formula

A240063(n) + a(n) + A240065(n) = A000041(n) for n >= 0.
G.f.: P(x)*(1 - x^2)*(1 - x^3)/(1 - x^5) where P(x) is the g.f. of A000041. - Andrew Howroyd, Jan 01 2025

A240065 Number of partitions of n such that m(2) > m(3), where m = multiplicity.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 4, 6, 9, 12, 17, 23, 33, 43, 59, 76, 102, 132, 173, 221, 285, 361, 462, 580, 733, 913, 1144, 1418, 1761, 2168, 2673, 3273, 4012, 4885, 5952, 7212, 8741, 10546, 12719, 15279, 18344, 21949, 26247, 31287, 37268, 44267, 52541, 62207, 73584
Offset: 0

Views

Author

Clark Kimberling, Mar 31 2014

Keywords

Examples

			a(6) counts these 4 partitions:  42, 222, 2211, 21111.
		

Crossrefs

Programs

  • Mathematica
    z = 60; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Count[p, 2] < Count[p, 3]], {n, 0, z}]  (* A240063 *)
    t2 = Table[Count[f[n], p_ /; Count[p, 2] <= Count[p, 3]], {n, 0, z}] (* A240063(n+3) *)
    t3 = Table[Count[f[n], p_ /; Count[p, 2] == Count[p, 3]], {n, 0, z}] (* A240064 *)
    t4 = Table[Count[f[n], p_ /; Count[p, 2] > Count[p, 3]], {n, 0, z}]  (* A240065 *)
    t5 = Table[Count[f[n], p_ /; Count[p, 2] >= Count[p, 3]], {n, 0, z}] (* A240065(n+2) *)

Formula

a(n) + A240063(n) + A240064(n) = A000041(n) for n >= 0.

A194703 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (3 + m).

Original entry on oeis.org

3, 2, 1, 0, 1, 2, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 3, 2, 1, 0, 1, 2, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Omar E. Pol, Feb 05 2012

Keywords

Comments

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 3. For further information see A182703 and A135010.

Examples

			Triangle begins:
3,
2, 1,
0, 1, 2,
1, 0, 1, 1,
0, 1, 0, 1, 1,
0, 0, 1, 0, 1, 1,
0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 0, 0, 1, 0, 1, 1,
0, 0, 0, 0, 0, 0, 1, 0, 1, 1,
...
For k = 1 and m = 1, T(1,1) = 3 because there are three parts of size 1 in the last section of the set of partitions of 4, since 3 + m = 4, so a(1) = 3.
For k = 2 and m = 1, T(2,1) = 2 because there are two parts of size 2 in the last section of the set of partitions of 4, since 3 + m = 4, so a(2) = 2.
		

Crossrefs

Always the sum of row k = p(3) = A000041(3) = 3.
The first (0-10) members of this family of triangles are A023531, A129186, A194702, this sequence, A194704-A194710.

Formula

T(k,m) = A182703(3+m,k), with T(k,m) = 0 if k > 3+m.
T(k,m) = A194812(3+m,k).

A206556 Number of 6's in the last section of the set of partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 5, 4, 8, 9, 14, 16, 26, 28, 42, 50, 69, 82, 114, 133, 179, 215, 279, 335, 434, 516, 657, 789, 987, 1182, 1473, 1754, 2164, 2583, 3154, 3755, 4567, 5414, 6542, 7753, 9307, 11000, 13158, 15501, 18456, 21712, 25731, 30196, 35677
Offset: 1

Views

Author

Omar E. Pol, Feb 09 2012

Keywords

Comments

Zero together with the first differences of A024790. Also number of occurrences of 6 in all partitions of n that do not contain 1 as a part. For the definition of "last section of n" see A135010. It appears that the sums of six successive terms give the partition numbers A000041.

Crossrefs

Programs

  • Sage
    A206556 = lambda n: sum(list(p).count(6) for p in Partitions(n) if 1 not in p)

Formula

It appears that A000041(n) = Sum_{j=1..6} a(n+j), n >= 0.

A206557 Number of 7's in the last section of the set of partitions of n.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 5, 7, 9, 13, 16, 23, 28, 39, 48, 64, 79, 104, 128, 165, 204, 258, 317, 399, 487, 606, 739, 912, 1105, 1356, 1637, 1994, 2400, 2906, 3485, 4199, 5016, 6015, 7164, 8553, 10151, 12076, 14286, 16930, 19974, 23588, 27749
Offset: 1

Views

Author

Omar E. Pol, Feb 09 2012

Keywords

Comments

Zero together with the first differences of A024791. Also number of occurrences of 7 in all partitions of n that do not contain 1 as a part. For the definition of "last section of n" see A135010. It appears that the sums of seven successive terms give the partition numbers A000041.

Crossrefs

Programs

  • Sage
    A206557 = lambda n: sum(list(p).count(7) for p in Partitions(n) if 1 not in p)

Formula

It appears that A000041(n) = Sum_{j=1..7} a(n+j), n >= 0.

A207377 Triangle read by rows in which row n lists the parts of the last section of the set of partitions of n in nondecreasing order.

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 1, 2, 2, 4, 1, 1, 1, 1, 1, 2, 3, 5, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 6, 8
Offset: 1

Views

Author

Omar E. Pol, Feb 23 2012

Keywords

Comments

Starting from the first row; it appears that the total numbers of occurrences of k in k successive rows give the sequence A000041. For more information see A182703.

Examples

			Written as a triangle:
1;
1,2;
1,1,3;
1,1,1,2,2,4;
1,1,1,1,1,2,3,5;
1,1,1,1,1,1,1,2,2,2,2,3,3,4,6;
1,1,1,1,1,1,1,1,1,1,1,2,2,2,3,3,4,5,7;
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,4,4,4,5,6,8;
		

Crossrefs

Triangle similar to A135010. Mirror of A207378. Row n has length A138137(n). Row sums give A138879. Right border is A000027.
Previous Showing 11-20 of 21 results. Next