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A239931 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n-3).

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%I A239931 #36 Dec 07 2016 11:07:51
%S A239931 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,
%T A239931 32,23,25,7,25,27,27,29,11,11,29,31,31,33,9,9,33,35,13,13,35,37,37,39,
%U A239931 18,39,41,15,9,15,41,43,11,11,43,45,45,47,17,17,47,49,49,51,51,53,43,43,53,55,55,57,57,59,21,22,21,59,61,11,61,63,15,15,63
%N A239931 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n-3).
%C A239931 Row n is a palindromic composition of sigma(4n-3).
%C A239931 Row n is also the row 4n-3 of A237270.
%C A239931 Row n has length A237271(4n-3).
%C A239931 Row sums give A112610.
%C A239931 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.
%C A239931 For the parts of the symmetric representation of sigma(4n-2), see A239932.
%C A239931 For the parts of the symmetric representation of sigma(4n-1), see A239933.
%C A239931 For the parts of the symmetric representation of sigma(4n), see A239934.
%C A239931 We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - _Omar E. Pol_, Dec 06 2016
%e A239931 The irregular triangle begins:
%e A239931    1;
%e A239931    3,  3;
%e A239931    5,  3,  5;
%e A239931    7,  7;
%e A239931    9,  9;
%e A239931   11,  5,  5, 11;
%e A239931   13,  5, 13;
%e A239931   15, 15;
%e A239931   17,  7,  7, 17;
%e A239931   19, 19;
%e A239931   21, 21;
%e A239931   23, 32, 23;
%e A239931   25,  7, 25;
%e A239931   27, 27;
%e A239931   29, 11, 11, 29;
%e A239931   31, 31;
%e A239931   ...
%e A239931 Illustration of initial terms in the first quadrant of the spiral described in A239660:
%e A239931 .
%e A239931 .     _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 15
%e A239931 .    |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
%e A239931 .                                  |
%e A239931 .                                  |
%e A239931 .     _ _ _ _ _ _ _ _ _ _ _ _ _ 13 |
%e A239931 .    |_ _ _ _ _ _ _ _ _ _ _ _ _|   |
%e A239931 .                              |   |_ _ _
%e A239931 .                              |         |
%e A239931 .     _ _ _ _ _ _ _ _ _ _ _ 11 |         |_
%e A239931 .    |_ _ _ _ _ _ _ _ _ _ _|   |_ _ _      |_
%e A239931 .                          |         |_ _ 5  |_
%e A239931 .                          |         |_  |_    |_ _
%e A239931 .     _ _ _ _ _ _ _ _ _ 9  |_ _ _      |_  |       |
%e A239931 .    |_ _ _ _ _ _ _ _ _|   |_ _  |_ 5    |_|_      |
%e A239931 .                      |       |_ _|_ 5      |     |_ _ _ _ _ _ 15
%e A239931 .                      |           | |_      |               | |
%e A239931 .     _ _ _ _ _ _ _ 7  |_ _        |_  |     |_ _ _ _ _ 13   | |
%e A239931 .    |_ _ _ _ _ _ _|       |_        | |             | |     | |
%e A239931 .                  |         |_      |_|_ _ _ _ 11   | |     | |
%e A239931 .                  |_ _        |             | |     | |     | |
%e A239931 .     _ _ _ _ _ 5      |_      |_ _ _ _ 9    | |     | |     | |
%e A239931 .    |_ _ _ _ _|         |           | |     | |     | |     | |
%e A239931 .              |_ _ 3    |_ _ _ 7    | |     | |     | |     | |
%e A239931 .              |_  |         | |     | |     | |     | |     | |
%e A239931 .     _ _ _ 3    |_|_ _ 5    | |     | |     | |     | |     | |
%e A239931 .    |_ _ _|         | |     | |     | |     | |     | |     | |
%e A239931 .          |_ _ 3    | |     | |     | |     | |     | |     | |
%e A239931 .            | |     | |     | |     | |     | |     | |     | |
%e A239931 .     _ 1    | |     | |     | |     | |     | |     | |     | |
%e A239931 .    |_|     |_|     |_|     |_|     |_|     |_|     |_|     |_|
%e A239931 .
%e A239931 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].
%e A239931 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.
%Y A239931 Cf. A000203, A112610, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239660, A239932-A239934, A244050, A245092, A262626.
%K A239931 nonn,tabf
%O A239931 1,2
%A A239931 _Omar E. Pol_, Mar 29 2014