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%I #58 Aug 04 2022 05:16:22
%S 1,1,2,1,2,1,1,2,0,1,0,2,1,2,0,2,0,1,1,2,0,2,0,2,1,1,1,0,0,2,0,0,1,2,
%T 0,2,0,0,1,0,1,2,2,0,2,0,0,2,0,0,1,1,0,2,0,1,2,0,0,2,2,0,1,0,0,1,2,0,
%U 0,0,1,2,1,0,2,0,0,0,1,0,0,0,2,2,0,0,2,0,0,0,2,0,1,1,1,2,0,0,2,0,0,0
%N Irregular triangle read by rows: T(n,k) is the number of subparts of the symmetric representation of sigma(n) that arise from the (2*k-1)-th double-staircase of the double-staircases diagram of n described in A335616, n >= 1, k >= 1, and the first element of column k is in row A000384(k).
%C Conjecture 1: the number of nonzero terms in row n equals A082647(n).
%C Conjecture 2: column k lists positive integers interleaved with 2*k+2 zeros.
%C T(n,k) is also the number of staircases (or subparts) of the ziggurat diagram of n (described in A347186) that arise from the (2*k-1)-th double-staircase of the double-staircases diagram of n (described in A335616).
%C The k-th column of the triangle is related to the (2*k+1)-gonal numbers. For further information about this connection see A347186 and A347263.
%C Terms can be 0, 1 or 2.
%e Triangle begins:
%e -----------------------
%e n / k 1 2 3 4
%e -----------------------
%e 1 | 1;
%e 2 | 1;
%e 3 | 2;
%e 4 | 1;
%e 5 | 2;
%e 6 | 1, 1;
%e 7 | 2, 0;
%e 8 | 1, 0;
%e 9 | 2, 1;
%e 10 | 2, 0;
%e 11 | 2, 0;
%e 12 | 1, 1;
%e 13 | 2, 0;
%e 14 | 2, 0;
%e 15 | 2, 1, 1;
%e 16 | 1, 0, 0;
%e 17 | 2, 0, 0;
%e 18 | 1, 2, 0;
%e 19 | 2, 0, 0;
%e 20 | 1, 0, 1;
%e 21 | 2, 2, 0;
%e 22 | 2, 0, 0;
%e 23 | 2, 0, 0;
%e 24 | 1, 1, 0;
%e 25 | 2, 0, 1;
%e 26 | 2, 0, 0;
%e 27 | 2, 2, 0;
%e 28 | 1, 0, 0, 1;
%e ...
%e For n = 15 the calculation of the 15th row of triangle (in accordance with the geometric algorithm described in A347186) is as follows:
%e Stage 1 (Construction):
%e We draw the diagram called "double-staircases" with 15 levels described in A335616.
%e Then we label the five double-staircases (j = 1..5) as shown below:
%e _
%e _| |_
%e _| _ |_
%e _| | | |_
%e _| _| |_ |_
%e _| | _ | |_
%e _| _| | | |_ |_
%e _| | | | | |_
%e _| _| _| |_ |_ |_
%e _| | | _ | | |_
%e _| _| | | | | |_ |_
%e _| | _| | | |_ | |_
%e _| _| | | | | |_ |_
%e _| | | _| |_ | | |_
%e _| _| _| | _ | |_ |_ |_
%e |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _|
%e 1 2 3 4 5
%e .
%e Stage 2 (Debugging):
%e We remove the fourth double-staircase as it does not have at least one step at level 1 of the diagram starting from the base, as shown below:
%e _
%e _| |_
%e _| _ |_
%e _| | | |_
%e _| _| |_ |_
%e _| | _ | |_
%e _| _| | | |_ |_
%e _| | | | | |_
%e _| _| _| |_ |_ |_
%e _| | | | | |_
%e _| _| | | |_ |_
%e _| | _| |_ | |_
%e _| _| | | |_ |_
%e _| | | | | |_
%e _| _| _| _ |_ |_ |_
%e |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
%e 1 2 3 5
%e .
%e Stage 3 (Annihilation):
%e We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase.
%e The new diagram has two double-staircases and two simple-staircases as shown below:
%e _
%e | |
%e _ | | _
%e _| | _| |_ | |_
%e _| | | | | |_
%e _| | | | | |_
%e _| | _| |_ | |_
%e _| | | | | |_
%e _| | | | | |_
%e _| | _| _ |_ | |_
%e |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _|
%e 1 3 5
%e .
%e The diagram is called "ziggurat of 15".
%e The staircase labeled 1 arises from the double-staircase labeled 1 in the double-staircases diagram of 15. There is a pair of these staircases, so T(15,1) = 2, since the symmetric representation of sigma(15) is also the base of the three dimensional version of the ziggurat .
%e The double-staircase labeled 3 is the same in both diagrams, so T(15,2) = 1.
%e The double-staircase labeled 5 is the same in both diagrams, so T(15,3) = 1.
%e Therefore the 15th row of the triangle is [2, 1, 1].
%e The top view of the 3D-Ziggurat of 15 and the symmetric representation of sigma(15) with subparts look like this:
%e _ _
%e |_| | |
%e |_| | |
%e |_| | |
%e |_| | |
%e |_| | |
%e |_| | |
%e |_| | |
%e _ _ _|_| _ _ _|_|
%e _ _|_| 36 _ _| | 8
%e |_|_|_| | _ _|
%e _|_|_| _| |_|
%e |_|_| 1 |_ _| 1
%e | 34 | 7
%e _ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _|
%e |_|_|_|_|_|_|_|_| |_ _ _ _ _ _ _ _|
%e 36 8
%e .
%e Top view of the 3D-Ziggurat. The symmetric representation of
%e The ziggurat is formed by 3 of sigma(15) is formed by 3 parts.
%e polycubes with 107 cubes It has 4 subparts with 24 cells in
%e in total. It has 4 staircases total. It is the base of the ziggurat.
%e with 24 steps in total.
%e .
%Y Another (and more regular) version of A279387 and of A280940.
%Y Row sums give A001227.
%Y Row n has length A351846(n).
%Y Cf. A000384, A082647, A196020, A235791, A236104, A237048, A237591, A237593, A262626, A335616, A346875, A347186, A347263, A347529.
%K nonn,tabf
%O 1,3
%A _Omar E. Pol_, Feb 20 2022