OFFSET
1,1
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
EXAMPLE
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
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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
KEYWORD
nonn,tabf,more
AUTHOR
Omar E. Pol, Mar 29 2014
STATUS
approved