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%I A351819 #58 Aug 04 2022 05:16:22 %S A351819 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 A351819 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 A351819 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 A351819 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 A351819 Conjecture 1: the number of nonzero terms in row n equals A082647(n). %C A351819 Conjecture 2: column k lists positive integers interleaved with 2*k+2 zeros. %C A351819 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 A351819 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 A351819 Terms can be 0, 1 or 2. %e A351819 Triangle begins: %e A351819 ----------------------- %e A351819 n / k 1 2 3 4 %e A351819 ----------------------- %e A351819 1 | 1; %e A351819 2 | 1; %e A351819 3 | 2; %e A351819 4 | 1; %e A351819 5 | 2; %e A351819 6 | 1, 1; %e A351819 7 | 2, 0; %e A351819 8 | 1, 0; %e A351819 9 | 2, 1; %e A351819 10 | 2, 0; %e A351819 11 | 2, 0; %e A351819 12 | 1, 1; %e A351819 13 | 2, 0; %e A351819 14 | 2, 0; %e A351819 15 | 2, 1, 1; %e A351819 16 | 1, 0, 0; %e A351819 17 | 2, 0, 0; %e A351819 18 | 1, 2, 0; %e A351819 19 | 2, 0, 0; %e A351819 20 | 1, 0, 1; %e A351819 21 | 2, 2, 0; %e A351819 22 | 2, 0, 0; %e A351819 23 | 2, 0, 0; %e A351819 24 | 1, 1, 0; %e A351819 25 | 2, 0, 1; %e A351819 26 | 2, 0, 0; %e A351819 27 | 2, 2, 0; %e A351819 28 | 1, 0, 0, 1; %e A351819 ... %e A351819 For n = 15 the calculation of the 15th row of triangle (in accordance with the geometric algorithm described in A347186) is as follows: %e A351819 Stage 1 (Construction): %e A351819 We draw the diagram called "double-staircases" with 15 levels described in A335616. %e A351819 Then we label the five double-staircases (j = 1..5) as shown below: %e A351819 _ %e A351819 _| |_ %e A351819 _| _ |_ %e A351819 _| | | |_ %e A351819 _| _| |_ |_ %e A351819 _| | _ | |_ %e A351819 _| _| | | |_ |_ %e A351819 _| | | | | |_ %e A351819 _| _| _| |_ |_ |_ %e A351819 _| | | _ | | |_ %e A351819 _| _| | | | | |_ |_ %e A351819 _| | _| | | |_ | |_ %e A351819 _| _| | | | | |_ |_ %e A351819 _| | | _| |_ | | |_ %e A351819 _| _| _| | _ | |_ |_ |_ %e A351819 |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _| %e A351819 1 2 3 4 5 %e A351819 . %e A351819 Stage 2 (Debugging): %e A351819 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 A351819 _ %e A351819 _| |_ %e A351819 _| _ |_ %e A351819 _| | | |_ %e A351819 _| _| |_ |_ %e A351819 _| | _ | |_ %e A351819 _| _| | | |_ |_ %e A351819 _| | | | | |_ %e A351819 _| _| _| |_ |_ |_ %e A351819 _| | | | | |_ %e A351819 _| _| | | |_ |_ %e A351819 _| | _| |_ | |_ %e A351819 _| _| | | |_ |_ %e A351819 _| | | | | |_ %e A351819 _| _| _| _ |_ |_ |_ %e A351819 |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _| %e A351819 1 2 3 5 %e A351819 . %e A351819 Stage 3 (Annihilation): %e A351819 We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase. %e A351819 The new diagram has two double-staircases and two simple-staircases as shown below: %e A351819 _ %e A351819 | | %e A351819 _ | | _ %e A351819 _| | _| |_ | |_ %e A351819 _| | | | | |_ %e A351819 _| | | | | |_ %e A351819 _| | _| |_ | |_ %e A351819 _| | | | | |_ %e A351819 _| | | | | |_ %e A351819 _| | _| _ |_ | |_ %e A351819 |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _| %e A351819 1 3 5 %e A351819 . %e A351819 The diagram is called "ziggurat of 15". %e A351819 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 A351819 The double-staircase labeled 3 is the same in both diagrams, so T(15,2) = 1. %e A351819 The double-staircase labeled 5 is the same in both diagrams, so T(15,3) = 1. %e A351819 Therefore the 15th row of the triangle is [2, 1, 1]. %e A351819 The top view of the 3D-Ziggurat of 15 and the symmetric representation of sigma(15) with subparts look like this: %e A351819 _ _ %e A351819 |_| | | %e A351819 |_| | | %e A351819 |_| | | %e A351819 |_| | | %e A351819 |_| | | %e A351819 |_| | | %e A351819 |_| | | %e A351819 _ _ _|_| _ _ _|_| %e A351819 _ _|_| 36 _ _| | 8 %e A351819 |_|_|_| | _ _| %e A351819 _|_|_| _| |_| %e A351819 |_|_| 1 |_ _| 1 %e A351819 | 34 | 7 %e A351819 _ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _| %e A351819 |_|_|_|_|_|_|_|_| |_ _ _ _ _ _ _ _| %e A351819 36 8 %e A351819 . %e A351819 Top view of the 3D-Ziggurat. The symmetric representation of %e A351819 The ziggurat is formed by 3 of sigma(15) is formed by 3 parts. %e A351819 polycubes with 107 cubes It has 4 subparts with 24 cells in %e A351819 in total. It has 4 staircases total. It is the base of the ziggurat. %e A351819 with 24 steps in total. %e A351819 . %Y A351819 Another (and more regular) version of A279387 and of A280940. %Y A351819 Row sums give A001227. %Y A351819 Row n has length A351846(n). %Y A351819 Cf. A000384, A082647, A196020, A235791, A236104, A237048, A237591, A237593, A262626, A335616, A346875, A347186, A347263, A347529. %K A351819 nonn,tabf %O A351819 1,3 %A A351819 _Omar E. Pol_, Feb 20 2022