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%I #48 Sep 21 2018 08:14:10
%S 7,15,28,31,42,60,56,63,91,90,42,42,124,49,49,120,168,127,63,63,195,
%T 70,70,186,224,180,84,84,252,217,210,280,248,105,105,360,112,112,255
%N Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n).
%C Row n is a palindromic composition of sigma(4n).
%C Row n is also the row 4n of A237270.
%C Row n has length A237271(4n).
%C Row sums give A193553.
%C First differs from A193553 at a(11).
%C 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.
%C For the parts of the symmetric representation of sigma(4n-3), see A239931.
%C For the parts of the symmetric representation of sigma(4n-2), see A239932.
%C For the parts of the symmetric representation of sigma(4n-1), see A239933.
%C We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - _Omar E. Pol_, Dec 06 2016
%e The irregular triangle begins:
%e 7;
%e 15;
%e 28;
%e 31;
%e 42;
%e 60;
%e 56;
%e 63;
%e 91;
%e 90;
%e 42, 42;
%e 124;
%e 49, 49;
%e 120;
%e 168;
%e ...
%e Illustration of initial terms in the fourth quadrant of the spiral described in A239660:
%e .
%e . 7 15 28 31 42 60 56 63
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%e .
%e 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.
%e 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.
%e 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].
%e 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.
%Y Cf. A000203, A193553, A196020, A236104, A235791, A237048, A237270, A237271, A237591, A237593, A239660, A239931, A239932, A239933, A244050, A245092, A262626.
%K nonn,tabf,more
%O 1,1
%A _Omar E. Pol_, Mar 29 2014
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