A379631 Irregular triangle read by rows: T(n,m), n >= 1, m >= 1, in which row n lists the largest parts of the partitions of n into consecutive parts followed by the conjugate corresponding odd divisors of n in accordance with the theorem described in A379630.

1, 1, 2, 1, 3, 2, 3, 1, 4, 1, 5, 3, 5, 1, 6, 3, 3, 1, 7, 4, 7, 1, 8, 1, 9, 5, 4, 3, 9, 1, 10, 4, 5, 1, 11, 6, 11, 1, 12, 5, 3, 1, 13, 7, 13, 1, 14, 5, 7, 1, 15, 8, 6, 5, 5, 3, 15, 1, 16, 1, 17, 9, 17, 1, 18, 7, 6, 9, 3, 1, 19, 10, 19, 1, 20, 6, 5, 1, 21, 11, 8, 6, 7, 3, 21, 1, 22, 7, 11, 1, 23, 12, 23, 1, 24, 9, 3, 1

Offset

1,3

Comments

Consider that the mentioned partitions are ordered by increasing number of parts.

Row n gives the n-th row of A379633 together with the n-th row of A379634.

Links

Table of n, a(n) for n=1..98.

Example

Triangle begins:
   1,  1;
   2,  1;
   3,  2,  3,  1;
   4,  1;
   5,  3,  5,  1;
   6,  3,  3,  1;
   7,  4,  7,  1;
   8,  1;
   9,  5,  4,  3,  9,  1,
  10,  4,  5,  1;
  11,  6, 11,  1;
  12,  5,  3,  1;
  13,  7, 13,  1;
  14,  5,  7,  1;
  15,  8,  6,  5,  5,  3, 15,  1;
  16,  1;
  17,  9, 17,  1;
  18,  7,  6,  9,  3,  1;
  19, 10, 19,  1;
  20,  6,  5,  1;
  21, 11,  8,  6,  7,  3, 21,  1;
  ...
For n = 21 the partitions of 21 into consecutive parts are [21], [11, 10], [8, 7, 6], [6, 5, 4, 3, 2, 1].
On the other hand the odd divisors of 21 are [1, 3, 7, 21].
To determine how these partitions are related to the odd divisors we follow the two rules of the theorem described in A379630 as shown below:
The first partition is [21] and the number of parts is 1 and 1 is odd so the corresponding odd divisor of 21 is 1.
The second partition is [11, 10] and the number of parts is 2 and 2 is even so the corresponding odd divisor of 21 is equal to 11 + 10 = 21.
The third partition is [8, 7, 6] and the number of parts is 3 and 3 is odd so the corresponding odd divisor of 21 is 3.
The fourth partition is [6, 5, 4, 3, 2, 1] and the number of parts is 6 and 6 is even so the corresponding odd divisor of 21 is equal to 6 + 1 = 5 + 2 = 4 + 3 = 7.
Summarizing in a table:
  --------------------------------------
              Correspondence
  --------------------------------------
    Partitions of 21              Odd
    into consecutive           divisors
         parts                   of 21
  -------------------         ----------
   [21]   ....................     1
   [11, 10]   ................    21
   [8, 7, 6]  ................     3
   [6, 5, 4, 3, 2, 1]  .......     7
.
Then we can make a table of conjugate correspondence in which the four partitions are arrenged in four columns with the largest parts at the top as shown below:
  ------------------------------------------
           Conjugate correspondence
  ------------------------------------------
    Partitions of 21              Odd
    into consecutive           divisors
    parts as columns             of 21
  -------------------     ------------------
   21   11    8    6       7    3   21    1
    |   10    7    5       |    |    |    |
    |    |    6    4       |    |    |    |
    |    |    |    3       |    |    |    |
    |    |    |    2       |    |    |    |
    |    |    |    1       |    |    |    |
    |    |    |    |_______|    |    |    |
    |    |    |_________________|    |    |
    |    |___________________________|    |
    |_____________________________________|
.
Then removing all rows except the first row we have a table of conjugate correspondence for largest parts and odd divisors as shown below:
  -------------------     ------------------
     Largest parts           Odd divisors
  -------------------     ------------------
   21   11    8    6       7    3   21    1
    |    |    |    |_______|    |    |    |
    |    |    |_________________|    |    |
    |    |___________________________|    |
    |_____________________________________|
.
So the 21st row of the triangle is [21, 11, 8, 6, 7, 3, 21, 1].
.
Illustration of initial terms in an isosceles triangle demonstrating the theorem described in A379630:
.                                          _ _
                                         _|1|1|_
                                       _|2 _|_ 1|_
                                     _|3  |2|3|  1|_
                                   _|4   _| | |_   1|_
                                 _|5    |3 _|_ 5|    1|_
                               _|6     _| |3|3| |_     1|_
                             _|7      |4  | | |  7|      1|_
                           _|8       _|  _| | |_  |_       1|_
                         _|9        |5  |4 _|_ 3|  9|        1|_
                       _|10        _|   | |4|5| |   |_         1|_
                     _|11         |6   _| | | | |_  11|          1|_
                   _|12          _|   |5  | | |  3|   |_           1|_
                 _|13           |7    |  _| | |_  |   13|            1|_
               _|14            _|    _| |5 _|_ 7| |_    |_             1|_
             _|15             |8    |6  | |5|5| |  3|   15|              1|_
           _|16              _|     |   | | | | |   |     |_               1|_
         _|17               |9     _|  _| | | | |_  |_    17|                1|_
       _|18                _|     |7  |6  | | |  9|  3|     |_                 1|_
     _|19                 |10     |   |  _| | |_  |   |     19|                  1|_
   _|20                  _|      _|   | |6 _|_ 5| |   |_      |_                   1|_
  |21                   |11     |8    | | |6|7| | |    3|     21|                    1|
.
The geometrical structure of the above isosceles triangle is defined in A237593. See also the triangles A286000 and A379633.
Note that the diagram also can be interpreted as a template which after folding gives a 90 degree pop-up card which has essentially the same structure as the stepped pyramid described in A245092.
.

Crossrefs

Column 1 gives A000027.

Right border gives A000012.

The sum of row n equals A286015(n) + A000593(n).

The length of row n is A054844(n) = 2*A001227(n).

For another version with smallest parts see A379630.

The partitions of n into consecutive parts are in the n-th row of A299765. See also A286000.

The odd divisors of n are in the n-th row of A182469. See also A261697 and A261699.

Cf. A196020, A204217, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A245092, A262626, A379632, A379633, A379634.

Sequence in context: A349552 A278573 A108663 * A307314 A057940 A097285

Adjacent sequences: A379628 A379629 A379630 * A379632 A379633 A379634

Keyword

nonn,tabf

Author

Omar E. Pol, Dec 30 2024

Status

approved