%I #92 Sep 06 2023 11:49:22
%S 1,1,2,1,1,2,0,2,1,1,2,3,4,3,2,1,1,2,3,0,0,0,3,2,1,1,2,3,4,5,7,5,4,3,
%T 2,1,1,2,3,4,0,0,0,0,0,4,3,2,1,1,2,3,4,5,6,7,8,7,6,5,4,3,2,1,1,2,3,4,
%U 5,0,0,1,4,1,0,0,5,4,3,2,1,1,2,3,4,5,6,7,8,9,0
%N Irregular triangle read by rows: T(n,k) is the total number of cells with multiplicity in the k-th column of the ziggurat diagram of n.
%C The "ziggurat" diagram arises as a remnant of the double-staircases diagram described in A335616 after a geometric algorithm equivalent to the algorithm described in A280850 and A296508.
%C The geometric algorithm is also equivalent to the folding of the isosceles triangle described in A237593 forming the structure of the pyramid described in A245092.
%C The ziggurat diagram of n gives us an explanation about the parts, subparts and widths of the symmetric representation of sigma(n).
%C In the ziggurat diagram of n we have that:
%C The number of parts equals A237271(n).
%C The number of subparts equals A001227(n).
%C The number of steps in the central column equals A067742(n).
%C The total number of steps equals A000203(n).
%C The correspondence between both diagrams is because a three-dimensional version of the ziggurat of n can be constructed with units cubes and where the base of the structure is the symmetric representation of sigma(n).
%e Triangle begins:
%e 1;
%e 1, 2, 1;
%e 1, 2, 0, 2, 1;
%e 1, 2, 3, 4, 3, 2, 1;
%e 1, 2, 3, 0, 0, 0, 3, 2, 1;
%e 1, 2, 3, 4, 5, 7, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 0, 0, 0, 0, 0, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 0, 0, 1, 4, 1, 0, 0, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 5, 4, 3, 2, 1;
%e ...
%e Written as an isosceles triangle we can see the symmetry of every row as shown below:
%e 1;
%e 1, 2, 1;
%e 1, 2, 0, 2, 1;
%e 1, 2, 3, 4, 3, 2, 1;
%e 1, 2, 3, 0, 0, 0, 3, 2, 1;
%e 1, 2, 3, 4, 5, 7, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 0, 0, 0, 0, 0, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 0, 0, 1, 4, 1, 0, 0, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1;
%e 1, 2, 3, 4, 5, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 5, 4, 3, 2, 1;
%e ...
%e For n = 15 the ziggurat diagram of 15 looks like this:
%e _
%e | |
%e _ | | _
%e _| | _| |_ | |_
%e _| | | | | |_
%e _| | | | | |_
%e _| | _| |_ | |_
%e _| | | | | |_
%e _| | | | | |_
%e _| | _| _ |_ | |_
%e |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
%e 1 2 3 4 5 6 7 8 0 0 0 1 4 7 B 7 4 1 0 0 0 8 7 6 5 4 3 2 1
%e .
%e Where B = 10 + 1 = 11.
%e The left-hand part (or the left-hand staircase) has 8 steps.
%e The right-hand part (or the right-hand staircase) has 8 steps.
%e The central part (formed by two subparts or two staircases) has a total of 7 + 1 = 8 steps.
%e The number of parts equals A237271(15) = 3.
%e The number of subparts equals A001227(15) = 4.
%e The number of steps in the central column equals A067742(15) = 2.
%e The total number of steps equals A000203(15) = 24.
%e Compare the above diagram with the symmetric representation of sigma(15) with subparts as shown below:
%e _
%e | |
%e | |
%e | |
%e | |
%e | |
%e | |
%e | |
%e _ _ _|_|
%e _ _| | 8
%e | _ _|
%e _| |_|
%e |_ _| 1
%e | 7
%e _ _ _ _ _ _ _ _|
%e |_ _ _ _ _ _ _ _|
%e 8
%e .
%e The left-hand part has 8 square cells.
%e The right-hand part has 8 square cells.
%e The central part (formed by two subparts) has a total of 7 + 1 = 8 square cells.
%e The number of parts equals A237271(15) = 3.
%e The number of subparts equals A001227(15) = 4.
%e The number of square cells on the main diagonal equals A067742(15) = 2.
%e The total number of square cells equals A000203(15) = 24.
%Y Row lengths give A005408.
%Y Analog of A249351.
%Y Cf. A000203, A001227, A067742, A196020, A235791, A236104, A237270, A237271, A237591, A237593 (Dyck paths), A245092, A279387 (subparts), A280850 (algorithm), A280851, A296508, A335616.
%K nonn,tabf
%O 1,3
%A _Omar E. Pol_, Aug 29 2021