A182703 -id:A182703 - cp's OEIS frontend

A212011 Triangle read by rows: T(n,k) = sum of all parts of the last k shells of n.

1, 3, 4, 5, 8, 9, 11, 16, 19, 20, 15, 26, 31, 34, 35, 31, 46, 57, 62, 65, 66, 39, 70, 85, 96, 101, 104, 105, 71, 110, 141, 156, 167, 172, 175, 176, 94, 165, 204, 235, 250, 261, 266, 269, 270, 150, 244, 315, 354, 385, 400, 411, 416, 419, 420, 196, 346

Offset: 1

Omar E. Pol, Apr 26 2012

The set of partitions of n contains n shells (see A135010). It appears that the last k shells of n contain p(n-k) parts of size k, where p(n) = A000041(n). See also A182703.

			For n = 5 the illustration shows five sets containing the last k shells of 5 and below we can see that the sum of all parts of in each set:
--------------------------------------------------------
.  S{5}       S{4-5}     S{3-5}     S{2-5}     S{1-5}
--------------------------------------------------------
.  The        Last       Last       Last       The
.  last       two        three      four       five
.  shell      shells     shells     shells     shells
.  of 5       of 5       of 5       of 5       of 5
--------------------------------------------------------
.
.  5          5          5          5          5
.  3+2        3+2        3+2        3+2        3+2
.    1        4+1        4+1        4+1        4+1
.      1      2+2+1      2+2+1      2+2+1      2+2+1
.      1        1+1      3+1+1      3+1+1      3+1+1
.        1        1+1      1+1+1    2+1+1+1    2+1+1+1
.          1        1+1      1+1+1    1+1+1+1  1+1+1+1+1
. ---------- ---------- ---------- ---------- ----------
.     15         26         31         34         35
.
So row 5 lists 15, 26, 31, 34, 35.
.
Triangle begins:
1;
3,     4;
5,     8,   9;
11,   16,  19,  20;
15,   26,  31,  34,  35;
31,   46,  57,  62,  65,  66;
39,   70,  85,  96, 101, 104, 105;
71,  110, 141, 156, 167, 172, 175, 176;
94,  165, 204, 235, 250, 261, 266, 269, 270;
150, 244, 315, 354, 385, 400, 411, 416, 419, 420;

Mirror of triangle A212001. Column 1 is A138879. Right border is A066186.

Cf. A135010, A182703, A212000, A212010.

T(n,k) = A066186(n) - A066186(n-k).

T(n,k) = Sum_{j=n-k+1..n} A138879(j).

A194704 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (4 + m).

Original entry on oeis.org

5, 1, 4, 1, 2, 2, 0, 1, 1, 3, 1, 0, 1, 1, 2, 0, 1, 0, 1, 1, 2, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 4. For further information see A182703 and A135010.

			Triangle begins:
  5,
  1, 4,
  1, 2, 2,
  0, 1, 1, 3,
  1, 0, 1, 1, 2,
  ...
For k = 1 and m = 1: T(1,1) = 5 because there are five parts of size 1 in the last section of the set of partitions of 5, since 4 + m = 5, so a(1) = 5.
For k = 2 and m = 1: T(2,1) = 1 because there is only one part of size 2 in the last section of the set of partitions of 5, since 4 + m = 5, so a(2) = 1.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Always the sum of row k = p(4) = A000041(4) = 5.
The first (0-10) members of this family of triangles are A023531, A129186, A194702, A194703, this sequence, A194705-A194710.
Cf. A135010, A138121, A182712-A182714, A194812.

P(n)={my(M=matrix(n,n), d=4); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

T(k,m) = A182703(4+m,k), with T(k,m) = 0 if k > 4+m.

T(k,m) = A194812(4+m,k).

Terms a(16) and beyond from Andrew Howroyd, Feb 19 2020

A194705 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (5 + m).

Original entry on oeis.org

7, 4, 3, 2, 2, 3, 1, 1, 3, 2, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 2, 0, 1, 0, 1, 1, 2, 2, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 5. For further information see A182703 and A135010.

			Triangle begins:
  7,
  4, 3,
  2, 2, 3,
  1, 1, 3, 2,
  0, 1, 1, 2, 3,
  1, 0, 1, 1, 2, 2,
  0, 1, 0, 1, 1, 2, 2,
  ...
For k = 1 and m = 1: T(1,1) = 7 because there are seven parts of size 1 in the last section of the set of partitions of 6, since 5 + m = 6, so a(1) = 7.
For k = 2 and m = 1: T(2,1) = 4 because there are four parts of size 2 in the last section of the set of partitions of 6, since 5 + m = 6, so a(2) = 4.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Always the sum of row k = p(5) = A000041(5) = 7.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194704, this sequence, A194706-A194710.
Cf. A135010, A138121, A194812.

P(n)={my(M=matrix(n,n), d=5); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

T(k,m) = A182703(5+m,k), with T(k,m) = 0 if k > 5+m.

T(k,m) = A194812(5+m,k).

Terms a(29) and beyond from Andrew Howroyd, Feb 19 2020

A194706 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (6 + m).

Original entry on oeis.org

11, 3, 8, 2, 3, 6, 1, 3, 2, 5, 1, 1, 2, 3, 4, 0, 1, 1, 2, 2, 5, 1, 0, 1, 1, 2, 2, 4, 0, 1, 0, 1, 1, 2, 2, 4, 0, 0, 1, 0, 1, 1, 2, 2, 4, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 6. For further information see A182703 and A135010.

			Triangle begins:
  11,
   3, 8,
   2, 3, 6,
   1, 3, 2, 5,
   1, 1, 2, 3, 4,
   0, 1, 1, 2, 2, 5,
  ...
For k = 1 and m = 1: T(1,1) = 11 because there are 11 parts of size 1 in the last section of the set of partitions of 7, since 6 + m = 7, so a(1) = 11.
For k = 2 and m = 1: T(2,1) = 3 because there are three parts of size 2 in the last section of the set of partitions of 7, since 6 + m = 7, so a(2) = 3.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Always the sum of row k = p(6) = A000041(6) = 11.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194705, this sequence, A194707-A194710.
Cf. A135010, A138121, A194812.

P(n)={my(M=matrix(n,n), d=6); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

T(k,m) = A182703(6+m,k), with T(k,m) = 0 if k > 6+m.

T(k,m) = A194812(6+m,k).

Terms a(22) and beyond from Andrew Howroyd, Feb 19 2020

A194707 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (7 + m).

Original entry on oeis.org

15, 8, 7, 3, 6, 6, 3, 2, 5, 5, 1, 2, 3, 4, 5, 1, 1, 2, 2, 5, 4, 0, 1, 1, 2, 2, 4, 5, 1, 0, 1, 1, 2, 2, 4, 4, 0, 1, 0, 1, 1, 2, 2, 4, 4, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 7. For further information see A182703 and A135010.

			Triangle begins:
  15,
   8, 7,
   3, 6, 6,
   3, 2, 5, 5,
  ...
For k = 1 and m = 1: T(1,1) = 15 because there are 15 parts of size 1 in the last section of the set of partitions of 8, since 7 + m = 8, so a(1) = 15.
For k = 2 and m = 1: T(2,1) = 8 because there are eight parts of size 2 in the last section of the set of partitions of 8, since 7 + m = 8, so a(2) = 8.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Always the sum of row k = p(7) = A000041(7) = 15.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194706, this sequence, A194708-A194710.
Cf. A135010, A138121, A194812.

P(n)={my(M=matrix(n,n), d=7); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

T(k,m) = A182703(7+m,k), with T(k,m) = 0 if k > 7+m.

T(k,m) = A194812(7+m,k).

Terms a(11) and beyond from Andrew Howroyd, Feb 19 2020

A194708 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (8 + m).

Original entry on oeis.org

22, 7, 15, 6, 6, 10, 2, 5, 5, 10, 2, 3, 4, 5, 8, 1, 2, 2, 5, 4, 8, 1, 1, 2, 2, 4, 5, 7, 0, 1, 1, 2, 2, 4, 4, 8, 1, 0, 1, 1, 2, 2, 4, 4, 7, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 8. For further information see A182703 and A135010.

			Triangle begins:
  22,
   7, 15,
   6,  6, 10,
   2,  5,  5, 10,
   2,  3,  4,  5,  8,
   ...
For k = 1 and m = 1: T(1,1) = 22 because there are 22 parts of size 1 in the last section of the set of partitions of 9, since 8 + m = 9, so a(1) = 22.
For k = 2 and m = 1: T(2,1) = 7 because there are seven parts of size 2 in the last section of the set of partitions of 9, since 8 + m = 9, so a(2) = 7.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Always the sum of row k = p(8) = A000041(8) = 22.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194707, this sequence, A194709, A194710.
Cf. A135010, A138121, A194812.

P(n)={my(M=matrix(n,n), d=8); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

T(k,m) = A182703(8+m,k), with T(k,m) = 0 if k > 8+m.

T(k,m) = A194812(8+m,k).

Terms a(11) and beyond from Andrew Howroyd, Feb 19 2020

A194709 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (9 + m).

Original entry on oeis.org

30, 15, 15, 6, 10, 14, 5, 5, 10, 10, 3, 4, 5, 8, 10, 2, 2, 5, 4, 8, 9, 1, 2, 2, 4, 5, 7, 9, 1, 1, 2, 2, 4, 4, 8, 8, 0, 1, 1, 2, 2, 4, 4, 7, 9, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 4, 7, 8

Offset: 1

Omar E. Pol, Feb 05 2012

Sub-triangle of A182703 and also of A194812. Note that the sum of every row is also the number of partitions of 9. For further information see A182703 and A135010.

			Triangle begins:
  30;
  15, 15;
   6, 10, 14;
   5,  5, 10, 10;
   3,  4,  5,  8, 10;
   2,  2,  5,  4,  8, 9;
  ...
For k = 1 and  m = 1; T(1,1) = 30 because there are 30 parts of size 1 in the last section of the set of partitions of 10, since 9 + m = 10, so a(1) = 30. For k = 2 and m = 1; T(2,1) = 15 because there are 15 parts of size 2 in the last section of the set of partitions of 10, since 9 + m = 10, so a(2) = 15.

Andrew Howroyd, Table of n, a(n) for n = 1..1275 (rows 1..50)

Always the sum of row k = p(9) = A000041(n) = 30.
The first (0-10) members of this family of triangles are A023531, A129186, A194702-A194708, this sequence, A194710.
Cf. A135010, A138121, A194812.

P(n)={my(M=matrix(n,n), d=9); M[1,1]=numbpart(d); for(m=1, n, forpart(p=m+d, for(k=1, #p, my(t=p[k]); if(t<=n && m<=t, M[t, m]++)), [2, m+d])); M}
{ my(T=P(10)); for(n=1, #T, print(T[n, 1..n])) } \\ Andrew Howroyd, Feb 19 2020

T(k,m) = A182703(9+m,k), with T(k,m) = 0 if k > 9+m.

T(k,m) = A194812(9+m,k).

Terms a(7) and beyond from Andrew Howroyd, Feb 19 2020

A194795 Imbalance of the number of partitions of n.

Original entry on oeis.org

0, -1, 0, -2, 0, -4, 0, -7, 1, -11, 3, -18, 6, -28, 13, -42, 24, -64, 41, -96, 69, -141, 112, -208, 175, -303, 271, -437, 410, -629, 609, -898, 896, -1271, 1302, -1792, 1868, -2510, 2660, -3493, 3752, -4839, 5248, -6666, 7293, -9131, 10065, -12454

Offset: 1

Omar E. Pol, Feb 02 2012

sign

Consider the three-dimensional structure of the shell model of partitions version "tree". Note that only the parts > 1 produce the imbalance. The 1's are located in the central columns. Note that every column contains exactly the same parts, the same as a periodic table (see example). For more information see A135010.

			For n = 6 the illustration of the three views of the shell model with 6 shells shows an imbalance (see below):
------------------------------------------------------
Partitions                Tree             Table 1.0
of 6.                    A194805            A135010
------------------------------------------------------
6                   6                     6 . . . . .
3+3                   3                   3 . . 3 . .
4+2                     4                 4 . . . 2 .
2+2+2                     2               2 . 2 . 2 .
5+1                         1   5         5 . . . . 1
3+2+1                       1 3           3 . . 2 . 1
4+1+1                   4   1             4 . . . 1 1
2+2+1+1                   2 1             2 . 2 . 1 1
3+1+1+1                     1 3           3 . . 1 1 1
2+1+1+1+1                 2 1             2 . 1 1 1 1
1+1+1+1+1+1                 1             1 1 1 1 1 1
------------------------------------------------------
.
.                   6 3 4 2 1 3 5
.     Table 2.0     . . . . 1 . .     Table 2.1
.      A182982      . . . 2 1 . .      A182983
.                   . 3 . . 1 2 .
.                   . . 2 2 1 . .
.                   . . . . 1
------------------------------------------------------
The number of partitions with parts on the left hand side is equal to 7 and the number of partitions with parts on the right hand side is equal to 3, so a(6) = -7+3 = -4. On the other hand; for n = 6 the first n terms of A002865 (with positive indices) are 0, 1, 1, 2, 2, 4 therefore a(6) = 0-1+1-2+2-4 = -4.

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Cf. A000041, A002865, A135010, A182703, A187219, A194796, A194797, A194805, A194809.

with(combinat):
a:= proc(n) option remember;
      (-1)^n *(numbpart(n-1)-numbpart(n)) +`if`(n>1, a(n-1), 0)
    end:
seq(a(n), n=1..70); # Alois P. Heinz, Apr 09 2012

a[n_] := a[n] = (-1)^n*(PartitionsP[n-1]-PartitionsP[n]) + If[n>1, a[n-1], 0]; Table[a[n], {n, 1, 70}] (* Jean-François Alcover, Nov 11 2015, after Alois P. Heinz *)
nmax = 60; Rest[CoefficientList[Series[x/(1-x) - (1+x)/(1-x) * Product[1/((1 + x^(2*k-1))*(1 - x^(2*k))), {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Nov 11 2015 *)
nmax = 60; Rest[CoefficientList[Series[-x/(1+x) - (1-x)/(1+x) * Product[1/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Nov 11 2015 *)

a(n) = Sum_{k=1..n} (-1)^(k-1)*(p(k)-p(k-1)), where p(k) is the number of partitions of k.
a(n) = Sum_{k=1..n} (-1)^(k-1)*A002865(k).
a(n) = (-1)^(n+1) * (A240690(n+1) - A240690(n)) - 1. - Vaclav Kotesovec, Nov 11 2015
a(n) ~ (-1)^(n+1) * Pi * exp(Pi*sqrt(2*n/3)) / (24*sqrt(2)*n^(3/2)). - Vaclav Kotesovec, Nov 11 2015

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

Omar E. Pol, Feb 09 2012

nonn

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.

Column 8 of A182703 and of A194812.

Cf. A000041, A135010, A138121, A182712-A182714, A206555- A206557, A206559, A206560.

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

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

Omar E. Pol, Feb 09 2012

nonn

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.

Column 9 of A182703 and of A194812.

Cf. A000041, A135010, A138121, A182712-A182714, A206555-A206558, A206560.

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

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

cp's OEIS Frontend

A212011 Triangle read by rows: T(n,k) = sum of all parts of the last k shells of n.

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A194704 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (4 + m).

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A194705 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (5 + m).

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A194706 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (6 + m).

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A194707 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (7 + m).

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A194708 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (8 + m).

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A194709 Triangle read by rows: T(k,m) = number of occurrences of k in the last section of the set of partitions of (9 + m).

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