OFFSET
1,2
COMMENTS
Row n is a palindromic composition of sigma(4n-3).
Row n is also the row 4n-3 of A237270.
Row n has length A237271(4n-3).
Row sums give A112610.
Also row n lists the parts of the symmetric representation of sigma in the n-th arm of the first quadrant of the spiral described in A239660, see example.
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.
For the parts of the symmetric representation of sigma(4n), see A239934.
We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016
EXAMPLE
The irregular triangle begins:
1;
3, 3;
5, 3, 5;
7, 7;
9, 9;
11, 5, 5, 11;
13, 5, 13;
15, 15;
17, 7, 7, 17;
19, 19;
21, 21;
23, 32, 23;
25, 7, 25;
27, 27;
29, 11, 11, 29;
31, 31;
...
Illustration of initial terms in the first quadrant of the spiral described in A239660:
.
. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 15
. |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
. |
. |
. _ _ _ _ _ _ _ _ _ _ _ _ _ 13 |
. |_ _ _ _ _ _ _ _ _ _ _ _ _| |
. | |_ _ _
. | |
. _ _ _ _ _ _ _ _ _ _ _ 11 | |_
. |_ _ _ _ _ _ _ _ _ _ _| |_ _ _ |_
. | |_ _ 5 |_
. | |_ |_ |_ _
. _ _ _ _ _ _ _ _ _ 9 |_ _ _ |_ | |
. |_ _ _ _ _ _ _ _ _| |_ _ |_ 5 |_|_ |
. | |_ _|_ 5 | |_ _ _ _ _ _ 15
. | | |_ | | |
. _ _ _ _ _ _ _ 7 |_ _ |_ | |_ _ _ _ _ 13 | |
. |_ _ _ _ _ _ _| |_ | | | | | |
. | |_ |_|_ _ _ _ 11 | | | |
. |_ _ | | | | | | |
. _ _ _ _ _ 5 |_ |_ _ _ _ 9 | | | | | |
. |_ _ _ _ _| | | | | | | | | |
. |_ _ 3 |_ _ _ 7 | | | | | | | |
. |_ | | | | | | | | | | |
. _ _ _ 3 |_|_ _ 5 | | | | | | | | | |
. |_ _ _| | | | | | | | | | | | |
. |_ _ 3 | | | | | | | | | | | |
. | | | | | | | | | | | | | |
. _ 1 | | | | | | | | | | | | | |
. |_| |_| |_| |_| |_| |_| |_| |_|
.
For n = 7 we have that 4*7-3 = 25 and the 25th row of A237593 is [13, 5, 3, 1, 2, 1, 1, 2, 1, 3, 5, 13] and the 24th row of A237593 is [13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13] therefore between both Dyck paths there are three regions (or parts) of sizes [13, 5, 13], so row 7 is [13, 5, 13].
The sum of divisors of 25 is 1 + 5 + 25 = A000203(25) = 31. On the other hand the sum of the parts of the symmetric representation of sigma(25) is 13 + 5 + 13 = 31, equaling the sum of divisors of 25.
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Mar 29 2014
STATUS
approved