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 1-6 of 6 results.

A206437 Triangle read by rows: T(j,k) is the k-th part of the j-th region of the set of partitions of n, if 1 <= j <= A000041(n).

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Feb 14 2012

Keywords

Comments

Here the j-th "region" of the set of partitions of n (or more simply the j-th "region" of n) is defined to be the first h elements of the sequence formed by the smallest parts in nonincreasing order of the partitions of the largest part of the j-th partition of n, with the list of partitions in colexicographic order, where h = j - i, and i is the index of the previous partition of n whose largest part is greater than the largest part of the j-th partition of n, or i = 0 if such previous largest part does not exist. The largest part of the j-th region of n is A141285(j) and the number of parts is h = A194446(j).
Some properties of the regions of n:
- The number of regions of n equals the number of partitions of n (see A000041).
- The set of regions of n contain the sets of regions of all positive integers previous to n.
- The first j regions of n are also first j regions of all integers greater than n.
- The sums of all largest parts of all regions of n equals the total number of parts of all regions of n. See A006128(n).
- If T(j,1) is a record in the sequence then the leading diagonals of triangle formed by the first j rows give the partitions of n (see example).
- The rank of a region is the largest part minus the number of parts (see A194447).
- The sum of all ranks of the regions of n is equal to zero.
How to make a diagram of the regions and partitions of n: in the first quadrant of the square grid we draw a horizontal line {[0, 0],[n, 0]} of length n. Then we draw a vertical line {[0, 0],[0, p(n)]} of length p(n) where p(n) is the number of partitions of n. Then, for j = 1..p(n), we draw a horizontal line {[0, j],[g, j]} where g = A141285(j) is the largest part of the j-th partition of n, with the list of partitions in colexicographic order. Then, for n = 1 .. p(n), we draw a vertical line from the point [g,j] down to intercept the next segment in a lower row. So we have a number of closed regions. Then we divide each region of n in horizontal rectangles with shorter sides = 1. We can see that in the original rectangle of area n*p(n) each row contains a set of rectangles whose areas are equal to the parts of one of the partitions of n. Then each region of n is labeled according to the position of its largest part on axis "y". Note that each region of n is similar to a mirror version of the Young diagram of one of the partitions of s, where s is the sum of all parts of the region. See the illustrations of the seven regions of 5 in the Links section.
Note that if row j of triangle contains parts of size 1 then the parts of row j are the smallest parts of all partitions of T(j,1), (see A046746), and also T(j,1) is a record in the sequence and also j is the number of partitions of T(j,1), (see A000041). Otherwise, if row j does not contain parts of size 1 then the parts of row j are the emergent parts of the next record in the sequence (see A183152). Row j is also the partition of A186412(j).
Also triangle read by rows in which row r lists the parts of the last section of the set of partitions of r, ordered by regions, such that the previous parts to the part of size r are the emergent parts of the partitions of r (see A138152) and the rest are the smallest parts of the partitions of r (see example). - Omar E. Pol, Apr 28 2012

Examples

			-------------------------------------------
  Region j   Triangle of parts
-------------------------------------------
  1          1;
  2          2,1;
  3          3,1,1;
  4          2;
  5          4,2,1,1,1;
  6          3;
  7          5,2,1,1,1,1,1;
  8          2;
  9          4,2;
  10         3;
  11         6,3,2,2,1,1,1,1,1,1,1;
  12         3;
  13         5,2;
  14         4;
  15         7,3,2,2,1,1,1,1,1,1,1,1,1,1,1;
.
The rotated triangle shows each row as a partition:
                             7
                           4   3
                         5       2
                       3   2       2
                     6               1
                   3   3               1
                 4       2               1
               2   2       2               1
             5               1               1
           3   2               1               1
         4       1               1               1
       2   2       1               1               1
     3       1       1               1               1
   2   1       1       1               1               1
 1   1   1       1       1               1               1
.
Alternative interpretation of this sequence:
Triangle read by rows in which row r lists the parts of the last section of the set of partitions of r ordered by regions (see comments):
   [1];
   [2,1];
   [3,1,1];
   [2],[4,2,1,1,1];
   [3],[5,2,1,1,1,1,1];
   [2],[4,2],[3],[6,3,2,2,1,1,1,1,1,1,1];
   [3],[5,2],[4],[7,3,2,2,1,1,1,1,1,1,1,1,1,1,1];
		

Crossrefs

Positive integers in A193870. Column 1 is A141285. Row j has length A194446(j). Row sums give A186412. Records are A000027.

Programs

  • Mathematica
    lex[n_]:=DeleteCases[Sort@PadRight[Reverse /@ IntegerPartitions@n], x_ /; x==0, 2];
    reg = {}; l = {};
    For[j = 1, j <= 22, j++,
      mx = Max@lex[j][[j]]; AppendTo[l, mx];
      For[i = j, i > 0, i--, If[l[[i]] > mx, Break[]]];
      AppendTo[reg, Take[Reverse[First /@ lex[mx]], j - i]];
      ];
    Flatten@reg  (* Robert Price, Apr 21 2020, revised Jul 24 2020 *)

Extensions

Further edited by Omar E. Pol, Mar 31 2012, Jan 27 2013
Minor edits by Omar E. Pol, Apr 23 2020
Comments corrected (following a suggestion from Peter Munn) by Omar E. Pol, Jul 20 2025

A186114 Triangle of regions and partitions of integers (see Comments lines for definition).

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 0, 0, 0, 2, 1, 1, 1, 2, 4, 0, 0, 0, 0, 0, 3, 1, 1, 1, 1, 1, 2, 5, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 7
Offset: 1

Views

Author

Omar E. Pol, Aug 08 2011

Keywords

Comments

Let r = T(n,k) be a record in the sequence. The consecutive records "r" are the natural numbers A000027. Consider the first n rows; the triangle T(n,k) has the property that the columns, without the zeros, from k..1, are also the partitions of r in juxtaposed reverse-lexicographical order, so k is also A000041(r), the number of partitions of r. Note that a record r is always the final term of a row if such row contains 1’s. The number of positive integer a(1)..r is A006128(r). The sums a(1)..r is A066186(r). Here the set of positive integers in every row (from 1 to n) is called a “region” of r. The number of regions of r equals the number of partitions of r. If T(n,1) = 1 then the row n is formed by the smallest parts, in nondecreasing order, of all partitions of T(n,n).

Examples

			Triangle begins:
1,
1, 2,
1, 1, 3,
0, 0, 0, 2,
1, 1, 1, 2, 4,
0, 0, 0, 0, 0, 3,
1, 1, 1, 1, 1, 2, 5,
0, 0, 0, 0, 0, 0, 0, 2,
0, 0, 0, 0, 0, 0, 0, 2, 4,
0, 0, 0, 0, 0, 0, 0, 0, 0, 3,
1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 6
...
The row n = 11 contains the 6th record in the sequence:  a(66) = T(11,11) = 6, then consider the first 11 rows of triangle. Note that the columns, from k = 11..1, without the zeros, are also the 11 partitions of 6 in juxtaposed reverse-lexicographical order: [6], [3, 3], [4, 2], [2, 2, 2], [5, 1], [3, 2, 1], [4, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [2, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1]. See A026792.
		

Crossrefs

Mirror of triangle A193870. Column 1 gives A167392. Right diagonal gives A141285.

Programs

Formula

T(n,1) = A167392(n).
T(n,k) = A141285(n), if k = n.

A193870 Triangle of regions and partitions of integers (see Comments lines for definition).

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 2, 0, 0, 0, 4, 2, 1, 1, 1, 3, 0, 0, 0, 0, 0, 5, 2, 1, 1, 1, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0, 4, 2, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Omar E. Pol, Aug 07 2011

Keywords

Comments

Triangle T(n,k) read by rows in which, from rows 1..n, if r = T(n,k) is a record in the sequence then the set of positive integers in every row (from 1 to n) is called a “region” of r. Note that n, the number of regions of r is also the number of partitions of r. The consecutive records "r" are the natural numbers A000027. The triangle has the property that, for rows n..1, the diagonals (without the zeros) are also the partitions of r, in juxtaposed reverse-lexicographical order. Note that a record "r" is the initial term of a row if such row contains 1’s. If T(n,k) is a record in the sequence then A000041(T(n,k)) = n. Note that if T(n,k) < 2 is not the last term of the row n then T(n,k+1) = T(n,k). The union of the rows that contain 1's gives A182715.

Examples

			Triangle begins:
1,
2, 1,
3, 1, 1,
2, 0, 0, 0,
4, 2, 1, 1, 1,
3, 0, 0, 0, 0, 0,
5, 2, 1, 1, 1, 1, 1,
2, 0, 0, 0, 0, 0, 0, 0,
4, 2, 0, 0, 0, 0, 0, 0, 0,
3, 0, 0, 0, 0, 0, 0, 0, 0, 0,
6, 3, 2, 2, 1, 1, 1, 1, 1, 1, 1,
…
For n = 11 note that the row n contains the 6th record in the sequence: T(11,1) = a(56) = 6, then consider the first 11 rows of triangle. Note that the diagonals d, from d = n..1, without the zeros, are also the partitions of 6 in juxtaposed reverse-lexicographical order: [6], [3, 3], [4, 2], [2, 2, 2], [5, 1], [3, 2, 1], [4, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [2, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1]. See A026792.
		

Crossrefs

Mirror of triangle A186114. Column 1 gives A141285. Right diagonal gives A167392.

Programs

Formula

T(n,1) = A141285(n).
T(n,k) = A167392(n), if k = n.

A210942 Triangle read by rows in which row n lists the parts > 1 of the n-th region of the shell model of partitions, with a(1) = 1.

Original entry on oeis.org

1, 2, 3, 2, 4, 2, 3, 5, 2, 2, 4, 2, 3, 6, 3, 2, 2, 3, 5, 2, 4, 7, 3, 2, 2, 2, 4, 2, 3, 6, 3, 2, 2, 5, 4, 8, 4, 3, 2, 2, 2, 2, 3, 5, 2, 4, 7, 3, 2, 2, 3, 6, 3, 5, 9, 4, 3, 3, 2, 2, 2, 2, 2, 4, 2, 3, 6, 3, 2, 2, 5, 4, 8, 4, 3, 2, 2, 2, 2, 4, 7, 3, 6, 5, 10, 5
Offset: 1

Views

Author

Omar E. Pol, Apr 18 2012

Keywords

Comments

For the definition of "region of n" see A206437. See also A186114. Row n lists the largest part and the parts > 1 of the n-th region of the shell model of partitions. Also 1 together with the numbers > 1 of A206437.

Examples

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

Crossrefs

Column 1 is A141285. Records give A000027. The n-th record is T(A000041(n),1).

A182377 Total sum of positive ranks of all regions in the last shell of n.

Original entry on oeis.org

0, 0, 0, 1, 2, 5, 8, 14, 21, 32, 45, 67, 91
Offset: 1

Views

Author

Omar E. Pol, Apr 29 2012

Keywords

Comments

The rank of a region of n is the largest part minus the number of parts. For the definition of "region of n" see A206437. For the definition of "last shell of n" see A135010.
a(n) is also the sum of positive integers in row n of triangle A194447. First differs from A000094 at a(12).

Examples

			For n = 7 the last shell of 7 contains four regions: [3],[5,2],[4],[7,3,2,2,1,1,1,1,1,1,1,1,1,1,1] so we have:
----------------------------------------------------------
.        Largest    Number
Region     part    of parts    Rank
----------------------------------------------------------
.  1        3         1          2
.  2        5         2          3
.  3        4         1          3
.  4        7        15         -8
.
The sum of positive ranks is a(7) = 2 + 3 + 3 = 8.
		

Crossrefs

A194449 Largest part minus the number of parts > 1 in the n-th region of the set of partitions of j, if 1 <= n <= A000041(j).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 2, 2, 3, 3, 3, 1, 2, 2, 2, 4, 3, 1, 2, 3, 3, 3, 2, 4, 4, 1, 1, 2, 2, 2, 4, 3, 1, 3, 5, 5, 4, -2, 2, 3, 3, 3, 2, 4, 4, 1, 4, 3, 5, 6, 5, -3, 1, 2, 2, 2, 4, 3, 1, 3, 5, 5, 4, -2, 2, 4, 4, 5, 3, 6, 6, 5, -9
Offset: 1

Views

Author

Omar E. Pol, Dec 10 2011

Keywords

Comments

Also triangle read by rows: T(j,k) = largest part minus the numbers of parts > 1 in the k-th region of the last section of the set of partitions of j. It appears that the sum of row j is equal to A000041(j-1). For the definition of "region" of the set of partitions of j see A206437. See also A135010.

Examples

			The 7th region of the shell model of partitions is [5, 2, 1, 1, 1, 1, 1]. The largest part is 5 and the number of parts > 1 is 2, so a(7) = 5 - 2 = 3 (see an illustration in the link section).
Written as an irregular triangle T(j,k) begins:
1;
1;
2;
1,2;
2,3;
1,2,2,2;
2,3,3,3;
1,2,2,2,4,3,1;
2,3,3,3,2,4,4,1;
1,2,2,2,4,3,1,3,5,5,4,-2;
2,3,3,3,2,4,4,1,4,3,5,6,5,-3;
1,2,2,2,4,3,1,3,5,5,4,-2,2,4,4,5,3,6,6,5,-9;
		

Crossrefs

Formula

a(n) = A141285(n) - A194448(n).
Showing 1-6 of 6 results.