|
%I #36 Dec 07 2016 11:07:51
%S 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 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 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 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(4n-3).
%C Row n is a palindromic composition of sigma(4n-3).
%C Row n is also the row 4n-3 of A237270.
%C Row n has length A237271(4n-3).
%C Row sums give A112610.
%C 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 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 For the parts of the symmetric representation of sigma(4n), see A239934.
%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 1;
%e 3, 3;
%e 5, 3, 5;
%e 7, 7;
%e 9, 9;
%e 11, 5, 5, 11;
%e 13, 5, 13;
%e 15, 15;
%e 17, 7, 7, 17;
%e 19, 19;
%e 21, 21;
%e 23, 32, 23;
%e 25, 7, 25;
%e 27, 27;
%e 29, 11, 11, 29;
%e 31, 31;
%e ...
%e Illustration of initial terms in the first quadrant of the spiral described in A239660:
%e .
%e . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 15
%e . |_ _ _ _ _ _ _ _ _ _ _ _ _ _ _|
%e . |
%e . |
%e . _ _ _ _ _ _ _ _ _ _ _ _ _ 13 |
%e . |_ _ _ _ _ _ _ _ _ _ _ _ _| |
%e . | |_ _ _
%e . | |
%e . _ _ _ _ _ _ _ _ _ _ _ 11 | |_
%e . |_ _ _ _ _ _ _ _ _ _ _| |_ _ _ |_
%e . | |_ _ 5 |_
%e . | |_ |_ |_ _
%e . _ _ _ _ _ _ _ _ _ 9 |_ _ _ |_ | |
%e . |_ _ _ _ _ _ _ _ _| |_ _ |_ 5 |_|_ |
%e . | |_ _|_ 5 | |_ _ _ _ _ _ 15
%e . | | |_ | | |
%e . _ _ _ _ _ _ _ 7 |_ _ |_ | |_ _ _ _ _ 13 | |
%e . |_ _ _ _ _ _ _| |_ | | | | | |
%e . | |_ |_|_ _ _ _ 11 | | | |
%e . |_ _ | | | | | | |
%e . _ _ _ _ _ 5 |_ |_ _ _ _ 9 | | | | | |
%e . |_ _ _ _ _| | | | | | | | | |
%e . |_ _ 3 |_ _ _ 7 | | | | | | | |
%e . |_ | | | | | | | | | | |
%e . _ _ _ 3 |_|_ _ 5 | | | | | | | | | |
%e . |_ _ _| | | | | | | | | | | | |
%e . |_ _ 3 | | | | | | | | | | | |
%e . | | | | | | | | | | | | | |
%e . _ 1 | | | | | | | | | | | | | |
%e . |_| |_| |_| |_| |_| |_| |_| |_|
%e .
%e 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 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 Cf. A000203, A112610, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239660, A239932-A239934, A244050, A245092, A262626.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Mar 29 2014
|
