A239934 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n).

7, 15, 28, 31, 42, 60, 56, 63, 91, 90, 42, 42, 124, 49, 49, 120, 168, 127, 63, 63, 195, 70, 70, 186, 224, 180, 84, 84, 252, 217, 210, 280, 248, 105, 105, 360, 112, 112, 255

Offset: 1

Author

Omar E. Pol, Mar 29 2014

Keywords

• nonn
• tabf
• more

Comments

Row n is a palindromic composition of sigma(4n).

Row n is also the row 4n of A237270.

Row n has length A237271(4n).

Row sums give A193553.

First differs from A193553 at a(11).

Also row n lists the parts of the symmetric representation of sigma in the n-th arm of the fourth quadrant of the spiral described in A239660, see example.

For the parts of the symmetric representation of sigma(4n-3), see A239931.

For the parts of the symmetric representation of sigma(4n-2), see A239932.

For the parts of the symmetric representation of sigma(4n-1), see A239933.

We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016

Examples

			The irregular triangle begins:
    7;
   15;
   28;
   31;
   42;
   60;
   56;
   63;
   91;
   90;
   42, 42;
  124;
   49, 49;
  120;
  168;
  ...
Illustration of initial terms in the fourth quadrant of the spiral described in A239660:
.
.           7       15      28      31      42      60      56      63
.           _       _       _       _       _       _       _       _
.          | |     | |     | |     | |     | |     | |     | |     | |
.         _| |     | |     | |     | |     | |     | |     | |     | |
.     _ _|  _|     | |     | |     | |     | |     | |     | |     | |
.    |_ _ _|    _ _| |     | |     | |     | |     | |     | |     | |
.             _|  _ _|     | |     | |     | |     | |     | |     | |
.            |  _|    _ _ _| |     | |     | |     | |     | |     | |
.     _ _ _ _| |    _|    _ _|     | |     | |     | |     | |     | |
.    |_ _ _ _ _|  _|     |    _ _ _| |     | |     | |     | |     | |
.                |      _|   |  _ _ _|     | |     | |     | |     | |
.                |  _ _|    _| |    _ _ _ _| |     | |     | |     | |
.     _ _ _ _ _ _| |      _|  _|   |  _ _ _ _|     | |     | |     | |
.    |_ _ _ _ _ _ _|  _ _|  _|  _ _| |    _ _ _ _ _| |     | |     | |
.                    |  _ _|  _|    _|   |    _ _ _ _|     | |     | |
.                    | |     |     |  _ _|   |    _ _ _ _ _| |     | |
.     _ _ _ _ _ _ _ _| |  _ _|  _ _|_|       |   |  _ _ _ _ _|     | |
.    |_ _ _ _ _ _ _ _ _| |  _ _|  _|      _ _|   | |    _ _ _ _ _ _| |
.                        | |     |      _|    _ _| |   |  _ _ _ _ _ _|
.                        | |  _ _|    _|  _ _|  _ _|   | |
.     _ _ _ _ _ _ _ _ _ _| | |       |   |    _|    _ _| |
.    |_ _ _ _ _ _ _ _ _ _ _| |  _ _ _|  _|  _|     |  _ _|
.                            | |       |  _|      _| |
.                            | |  _ _ _| |      _|  _|
.     _ _ _ _ _ _ _ _ _ _ _ _| | |  _ _ _|  _ _|  _|
.    |_ _ _ _ _ _ _ _ _ _ _ _ _| | |       |  _ _|
.                                | |  _ _ _| |
.                                | | |  _ _ _|
.     _ _ _ _ _ _ _ _ _ _ _ _ _ _| | | |
.    |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _| | |
.                                    | |
.                                    | |
.     _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _| |
.    |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
.
For n = 7 we have that 4*7 = 28 and the 28th row of A237593 is [15, 5, 3, 2, 1, 1, 1, 1, 1, 1, 2, 3, 5, 15] and the 27th row of A237593 is [14, 5, 3, 2, 1, 2, 2, 1, 2, 3, 5, 14] therefore between both Dyck paths there are only one region (or part) of size 56, so row 7 is 56.
The sum of divisors of 28 is 1 + 2 + 4 + 7 + 14 + 28 = A000203(28) = 56. On the other hand the sum of the parts of the symmetric representation of sigma(28) is 56, equaling the sum of divisors of 28.
For n = 11 we have that 4*11 = 44 and the 44th row of A237593 is [23, 8, 4, 3, 2, 1, 1, 2, 2, 1, 1, 2, 3, 4, 8, 23] and the 43rd row of A237593 is [22, 8, 4, 3, 2, 1, 2, 1, 1, 2, 1, 2, 3, 4, 8, 23] therefore between both Dyck paths there are two regions (or parts) of sizes [42, 42], so row 11 is [42, 42].
The sum of divisors of 44 is 1 + 2 + 4 + 11 + 22 + 44 = A000203(44) = 84. On the other hand the sum of the parts of the symmetric representation of sigma(44) is 42 + 42 = 84, equaling the sum of divisors of 44.
		

Crossrefs

Cf. A000203, A193553, A196020, A236104, A235791, A237048, A237270, A237271, A237591, A237593, A239660, A239931, A239932, A239933, A244050, A245092, A262626.