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%I A347186 #214 Jul 19 2024 12:04:49 %S A347186 1,4,6,16,12,37,20,64,36,90,42,161,56,156,107,256,90,334,110,408,202, %T A347186 342,156,697,207,462,312,785,240,976,272,1024,446,756,441,1586,380, %U A347186 930,604,1736,462,1841,506,1806,1101,1332,600,2921,720,1820,992,2450,756,2998,1108,3257 %N A347186 Ziggurat sequence (see Comments lines for definition). %C A347186 What do the parts and subparts of the symmetric representation of sigma(n) represent? (cf. A237270, A280851). This sequence gives an answer. %C A347186 To calculate a(n) we must follow a geometric algorithm composed of three stages as follows: %C A347186 Stage 1 (Construction): %C A347186 On the infinite square grid we draw the diagram called "double-staircases" with n levels described in A335616. %C A347186 Then we label the double-staircases from left to right starting from k = 1 to the central column of the diagram. %C A347186 Note that k is also the difference in height between two steps of the ladder it would have if we drew it with more than one step. %C A347186 Stage 2 (Debugging): %C A347186 We remove all double-staircases that do not have at least one step at the level 1 of the diagram starting from the base. %C A347186 Now the number of steps in the k-th double-staircase is equal to A196020(n,k). %C A347186 Stage 3 (Annihilation): %C A347186 From left to right, we remove each even-indexed double-staircase along with the steps of the nearest odd-indexed double-staircase that are just above it. %C A347186 As a result of this geometric algorithm a diagram is obtained which can have only double-staircases, only simple-staircases, or both at the same time. %C A347186 We call double-staircases those that have a step in the central column of the diagram. The rest are simple-staircases forming one or more symmetrical pairs equidistant from the central column of the diagram. %C A347186 We call "parts" of the diagram to the staircases (and the cells below them) that are separated from each other by columns of zero height. %C A347186 We call "subparts" of the diagram to the polygons formed by the cells that are under the staircases. %C A347186 a(n) is the total area (or the total number of cells) under all the staircases, with multiplicity. %C A347186 The connection with the polygonal numbers is as follows: %C A347186 The area under a double-staircase labeled with the number k is equal to the m-th (k+2)-gonal number plus the (m-1)-th (k+2)-gonal number, where m is the number of steps on one side of the ladder from the base to the top. %C A347186 If k = 1 then the area under the double-staircase is also equal to n^2 = A000290(n). %C A347186 The area under a simple-staircase labeled with the number k is equal to the m-th (k+2)-gonal number, where m is the number of steps. %C A347186 If n is a power of 2, or if n is an odd prime number, or if n is an even perfect number, or if n is a member of A246955, then the calculation of a(n) is easy (see the Formula section). %C A347186 The connection with the symmetric representation of sigma(n) or "SRS(n)" is as follows: %C A347186 The total number of steps in the diagram is equal to A000203(n), equaling the total area (or the number of cells) in the SRS(n). %C A347186 The number of parts in the diagram is equal to A237271(n) equaling the number of parts in the SRS(n). %C A347186 The number of double-staircases (also the number of steps in the central column in the diagram) is equal to A067742(n), equaling the number of central subparts in the SRS(n). %C A347186 The number of simple-staircases is equal to A281009(n), equaling the total number of equidistant subparts in the SRS(n). %C A347186 The total number of staircases is equal to A001227(n), equaling the number of subparts in the SRS(n). %C A347186 The number of columns in the diagram is equal to 2*n - 1, equaling the number of "widths" in the SRS(n) (cf. A249351). %C A347186 The number of steps in the successive parts of the diagram gives the n-th row of triangle A237270, equaling the successive parts in the SRS(n). %C A347186 The number of steps in the successive staircases from left to right gives the n-th row of triangle A280851, equaling the successive subparts from left to right in the SRS(n). %C A347186 The diagram is essentially the front view of a three-dimensional structure whose base is the symmetric representation of sigma(n), so a(n) is also the total number of cubic cells (or cubes) in the structure. %C A347186 The number of polycubes in the structure is equal to A237271(n), equaling the number of parts in the SRS(n). %C A347186 For some values of n the three-dimensional structure resembles a "ziggurat" which is a type of massive structure built in ancient Mesopotamia. %C A347186 It appears that the geometric algorithm described here is equivalent to the numerical algorithm conjectured in A280850 in the sense that both allow the value of the subparts of the SRS(n) to be computed. %C A347186 Both algorithms would also be equivalent to the geometric transformation of the isosceles triangle described in A237593 which, when folded, transforms into the vertical faces of the structure of the stepped pyramid described in A245092. %C A347186 Note that the stage 2 (Debugging) is in correspondence with the formula A196020(n,k) = A237048(n,k)*A338721(n,k) which transform the triangle A338721 into the triangle A196020. - _Omar E. Pol_, Jun 23 2022 %H A347186 Omar E. Pol, <a href="/A347186/a347186_1.jpg">Illustration of a(1)..a(8), 3D-Ziggurats</a> %H A347186 Omar E. Pol, <a href="/A347186/a347186_2.jpg">Illustration of a(9)..a(13), 3D-Ziggurats</a> %H A347186 Omar E. Pol, <a href="/A347186/a347186_3.jpg">Illustration of a(14)..a(15), 3D-Ziggurats</a> %F A347186 a(2^(n-1)) = n^2 = A000290(n). %F A347186 a(P) = 1 + P^2, if P is an even perfect number. %F A347186 a(p) = 2*A000217((p+1)/2) = A002378((p+1)/2), if p is an odd prime. %F A347186 a(k) = 2*A000217(A000203(k)/2), if k is a member of A246955. %e A347186 Illustration of the geometric algorithm and the initial terms (n = 1..6): %e A347186 ------------------------------------------------------------------------------- %e A347186 Stage 1 Stage 2 Stage 3 %e A347186 (Construction) (Debugging) (Annihilation) %e A347186 ------------------------------------------------------------------------------- %e A347186 Double-staircases Diagram of Ziggurat %e A347186 n diagram A196020 diagram a(n) %e A347186 ------------------------------------------------------------------------------- %e A347186 _ _ _ %e A347186 1 |_| |_| |_| 1 %e A347186 1 1 1 %e A347186 . %e A347186 _ _ _ %e A347186 _| |_ _| |_ _| |_ %e A347186 2 |_ _ _| |_ _ _| |_ _ _| 4 %e A347186 1 1 1 %e A347186 . %e A347186 _ _ %e A347186 _| |_ _| |_ _ _ %e A347186 _| _ |_ _| _ |_ _| | | |_ %e A347186 3 |_ _|_|_ _| |_ _|_|_ _| |_ _|_|_ _| 6 %e A347186 1 2 1 2 1 %e A347186 . %e A347186 _ _ _ %e A347186 _| |_ _| |_ _| |_ %e A347186 _| _ |_ _| |_ _| |_ %e A347186 _| | | |_ _| |_ _| |_ %e A347186 4 |_ _ _|_|_ _ _| |_ _ _ _ _ _ _| |_ _ _ _ _ _ _| 16 %e A347186 1 2 1 1 %e A347186 . %e A347186 _ _ %e A347186 _| |_ _| |_ %e A347186 _| _ |_ _| _ |_ _ _ %e A347186 _| | | |_ _| | | |_ _| | | |_ %e A347186 _| _| |_ |_ _| _| |_ |_ _| | | |_ %e A347186 5 |_ _ _|_ _ _|_ _ _| |_ _ _|_ _ _|_ _ _| |_ _ _|_ _ _|_ _ _| 12 %e A347186 1 2 1 2 1 %e A347186 . %e A347186 _ _ _ %e A347186 _| |_ _| |_ _| |_ %e A347186 _| _ |_ _| |_ _| |_ %e A347186 _| | | |_ _| |_ _| |_ %e A347186 _| _| |_ |_ _| |_ _| |_ %e A347186 _| | _ | |_ _| _ |_ _| _ |_ %e A347186 6 |_ _ _ _|_|_|_|_ _ _ _| |_ _ _ _ _|_|_ _ _ _ _| |_ _ _ _ _|_|_ _ _ _ _| 37 %e A347186 1 2 3 1 3 1 3 %e A347186 . %e A347186 For n = 7..14 the examples are omitted. %e A347186 For n = 15 the illustration of the geometric algorithm is as follows: %e A347186 Stage 1 (Construction): %e A347186 We draw the diagram called "double-staircases" with 15 levels described in A335616. %e A347186 Then we label the five double-staircases (k = 1..5) as shown below: %e A347186 _ %e A347186 _| |_ %e A347186 _| _ |_ %e A347186 _| | | |_ %e A347186 _| _| |_ |_ %e A347186 _| | _ | |_ %e A347186 _| _| | | |_ |_ %e A347186 _| | | | | |_ %e A347186 _| _| _| |_ |_ |_ %e A347186 _| | | _ | | |_ %e A347186 _| _| | | | | |_ |_ %e A347186 _| | _| | | |_ | |_ %e A347186 _| _| | | | | |_ |_ %e A347186 _| | | _| |_ | | |_ %e A347186 _| _| _| | _ | |_ |_ |_ %e A347186 |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _| %e A347186 1 2 3 4 5 %e A347186 . %e A347186 Stage 2 (Debugging): %e A347186 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 A347186 _ %e A347186 _| |_ %e A347186 _| _ |_ %e A347186 _| | | |_ %e A347186 _| _| |_ |_ %e A347186 _| | _ | |_ %e A347186 _| _| | | |_ |_ %e A347186 _| | | | | |_ %e A347186 _| _| _| |_ |_ |_ %e A347186 _| | | | | |_ %e A347186 _| _| | | |_ |_ %e A347186 _| | _| |_ | |_ %e A347186 _| _| | | |_ |_ %e A347186 _| | | | | |_ %e A347186 _| _| _| _ |_ |_ |_ %e A347186 |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _| %e A347186 1 2 3 5 %e A347186 . %e A347186 Note that the number of steps in the successive double-staircases gives [29, 13, 7, 0, 1], the same as the 15th row of triangle A196020 (whose alternate sums equals sigma(15) = A000203(15) = 24). %e A347186 Stage 3 (Annihilation): %e A347186 We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase. %e A347186 As a result of this geometric algorithm a new diagram is obtained which in this case has two double-staircases and two simple-staircases as shown below: %e A347186 _ %e A347186 | | %e A347186 _ | | _ %e A347186 _| | _| |_ | |_ %e A347186 _| | | | | |_ %e A347186 _| | | | | |_ %e A347186 _| | _| |_ | |_ %e A347186 _| | | | | |_ %e A347186 _| | | | | |_ %e A347186 _| | _| _ |_ | |_ %e A347186 |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _| %e A347186 1 3 5 %e A347186 . %e A347186 The diagram is called here "ziggurat of order 15". %e A347186 Now we calculate the total area (or the total number of cells) under the staircases with multiplicity using polygonal numbers as shown below: %e A347186 The area under the staircase labeled 1 is equal to A000217(8) = 36. There are a pair of this staircases, so the total area of this pair is equal to 2*36 = 72. %e A347186 The area under the double-staircase labeled 3 is equal to A000326(4) + A000326(3) = 22 + 12 = 34. %e A347186 The area under the double-staircase labeled 5 is equal to A000566(1) + A000566(0) = 1 + 0 = 1. %e A347186 Therefore the total area is a(15) = 72 + 34 + 1 = 107. %e A347186 The connection with the symmetric representation of sigma(15) or "SRS(15)" is as follows: %e A347186 The total number of steps is equal to A000203(15) = 24, equaling the total area (or number of cells) in the SRS(15). %e A347186 The number of parts in the diagram is equal to A237271(15) = 3 equaling the number of parts in the SRS(15). %e A347186 The number of double-staircases (also the number of steps in the central column in the diagram) is equal to A067742(15) = 2, equaling the number of central subparts in the SRS(15). %e A347186 The number of simple-staircases is equal to A281009(15) = 2, equaling the total number of equidistant subparts in the SRS(15). %e A347186 The total number of staircases is qual to A001227(15) = 4, equaling the number of subparts in the SRS(15). %e A347186 The number of columns in the diagram is equal to 2*15 - 1 = 29 equaling the number of "widths" in the SRS(15) (cf. A249351). %e A347186 The number of steps in the successive parts of the diagram are [8, 8, 8], the same as the 15th row of triangle A237270, matching the successive parts in the SRS(15). %e A347186 The number of steps in the successive staircases from left to right are respectively [8, 7, 1, 8], the same as the 15th row of triangle A280851, matching the successive subparts in the SRS(15). %e A347186 a(15) = 107 is also the number of cubic cells in the three-dimensional version of the structure whose base is the SRS(15). %e A347186 The number of polycubes in the structure is equal to A237271(15) = 3, equaling the number of parts in the SRS(15). %e A347186 The top view of the 3D-Ziggurat of order 15 and the symmetric representation of sigma(15) with subparts look like this: %e A347186 _ _ %e A347186 |_| | | %e A347186 |_| | | %e A347186 |_| | | %e A347186 |_| | | %e A347186 |_| | | %e A347186 |_| | | %e A347186 |_| | | %e A347186 _ _ _|_| _ _ _|_| %e A347186 _ _|_| 36 _ _| | 8 %e A347186 |_|_|_| | _ _| %e A347186 _|_|_| _| |_| %e A347186 |_|_| 1 |_ _| 1 %e A347186 | 34 | 7 %e A347186 _ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _| %e A347186 |_|_|_|_|_|_|_|_| |_ _ _ _ _ _ _ _| %e A347186 36 8 %e A347186 . %e A347186 Top view of the 3D-Ziggurat. The symmetric representation of %e A347186 The ziggurat is formed by 3 of sigma(15) is formed by 3 parts. %e A347186 polycubes with a(15) = 107 cubes It has 4 subparts with 24 cells in %e A347186 in total. It has 4 staircases total. It is the base of the ziggurat. %e A347186 with 24 steps in total. %e A347186 . %Y A347186 Row sums of A347367 and of A347263. %Y A347186 Cf. A356351 (partial sums). %Y A347186 Cf. A279387 (definition of subpart). %Y A347186 Cf. A000079, A000203, A000217, A000290, A000326, A000396, A001227, A002378, A065091, A067742, A131576, A196020, A235791, A236104, A237270, A237271, A237591, A237593, A245092, A246955, A249351, A262626, A280850, A280851, A281009, A296508, A296512, A296513, A335616, A338721, A347273, A347361, A347529, A351819. %K A347186 nonn %O A347186 1,2 %A A347186 _Omar E. Pol_, Aug 21 2021