This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A347529 #89 Jun 16 2023 03:17:00 %S A347529 1,3,4,7,6,11,1,8,0,15,0,10,3,18,0,12,0,23,5,14,0,24,0,16,7,1,31,0,0, %T A347529 18,0,0,35,4,0,20,0,0,39,0,3,22,10,0,36,0,0,24,0,0,47,13,0,26,0,5,42, %U A347529 0,0,28,12,0,55,0,0,1,30,0,0,0,59,6,7,0,32,0,0,0,63,0,0,0,34,14,0,0 %N A347529 Irregular triangle read by rows: T(n,k) is the sum of the 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 A347529 Conjecture 1: the number of nonzero terms in row n equals A082647(n). %C A347529 Conjecture 2: column k lists positive integers interleaved with 2*k+2 zeros. %C A347529 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 A347529 If n is prime then the only nonzero term in row n is T(n,1) = 1 + n. %C A347529 If n is a power of 2 then the only nonzero term in row n is T(n,1) = 2*n - 1. %C A347529 If n is an even perfect number then there are two nonzero terms in row n, they are T(n,1) = 2*n - 1 and the last term in the row is 1. %C A347529 If n is a hexagonal number then the last term in row n is 1. %C A347529 Row n contains a subpart 1 if and only if n is a hexagonal number. %C A347529 First differs from A279388 at a(10), or row 9 of triangle. %H A347529 <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a> %e A347529 Triangle begins: %e A347529 --------------------------- %e A347529 n / k 1 2 3 4 %e A347529 --------------------------- %e A347529 1 | 1; %e A347529 2 | 3; %e A347529 3 | 4; %e A347529 4 | 7; %e A347529 5 | 6; %e A347529 6 | 11, 1; %e A347529 7 | 8, 0; %e A347529 8 | 15, 0; %e A347529 9 | 10, 3; %e A347529 10 | 18, 0; %e A347529 11 | 12, 0; %e A347529 12 | 23, 5; %e A347529 13 | 14, 0; %e A347529 14 | 24, 0; %e A347529 15 | 16, 7, 1; %e A347529 16 | 31, 0, 0; %e A347529 17 | 18, 0, 0; %e A347529 18 | 35, 4, 0; %e A347529 19 | 20, 0, 0; %e A347529 20 | 39, 0, 3; %e A347529 21 | 22, 10, 0; %e A347529 22 | 36, 0, 0; %e A347529 23 | 24, 0, 0; %e A347529 24 | 47, 13, 0; %e A347529 25 | 26, 0, 5; %e A347529 26 | 42, 0, 0; %e A347529 27 | 28, 12, 0; %e A347529 28 | 55, 0, 0, 1; %e A347529 ... %e A347529 For n = 15 the calculation of the 15th row of the triangle (in accordance with the geometric algorithm described in A347186) is as follows: %e A347529 Stage 1 (Construction): %e A347529 We draw the diagram called "double-staircases" with 15 levels described in A335616. %e A347529 Then we label the five double-staircases (j = 1..5) as shown below: %e A347529 _ %e A347529 _| |_ %e A347529 _| _ |_ %e A347529 _| | | |_ %e A347529 _| _| |_ |_ %e A347529 _| | _ | |_ %e A347529 _| _| | | |_ |_ %e A347529 _| | | | | |_ %e A347529 _| _| _| |_ |_ |_ %e A347529 _| | | _ | | |_ %e A347529 _| _| | | | | |_ |_ %e A347529 _| | _| | | |_ | |_ %e A347529 _| _| | | | | |_ |_ %e A347529 _| | | _| |_ | | |_ %e A347529 _| _| _| | _ | |_ |_ |_ %e A347529 |_ _ _ _ _ _ _ _|_ _ _|_ _|_|_|_|_ _|_ _ _|_ _ _ _ _ _ _ _| %e A347529 1 2 3 4 5 %e A347529 . %e A347529 Stage 2 (Debugging): %e A347529 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 A347529 _ %e A347529 _| |_ %e A347529 _| _ |_ %e A347529 _| | | |_ %e A347529 _| _| |_ |_ %e A347529 _| | _ | |_ %e A347529 _| _| | | |_ |_ %e A347529 _| | | | | |_ %e A347529 _| _| _| |_ |_ |_ %e A347529 _| | | | | |_ %e A347529 _| _| | | |_ |_ %e A347529 _| | _| |_ | |_ %e A347529 _| _| | | |_ |_ %e A347529 _| | | | | |_ %e A347529 _| _| _| _ |_ |_ |_ %e A347529 |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _| %e A347529 1 2 3 5 %e A347529 . %e A347529 Stage 3 (Annihilation): %e A347529 We delete the second double-staircase and the steps of the first double-staircase that are just above the second double-staircase. %e A347529 The new diagram has two double-staircases and two simple-staircases as shown below: %e A347529 _ %e A347529 | | %e A347529 _ | | _ %e A347529 _| | _| |_ | |_ %e A347529 _| | | | | |_ %e A347529 _| | | | | |_ %e A347529 _| | _| |_ | |_ %e A347529 _| | | | | |_ %e A347529 _| | | | | |_ %e A347529 _| | _| _ |_ | |_ %e A347529 |_ _ _ _ _ _ _ _|_ _ _|_ _ _|_|_ _ _|_ _ _|_ _ _ _ _ _ _ _| %e A347529 1 3 5 %e A347529 . %e A347529 The diagram is called "ziggurat of 15". %e A347529 The number of steps in the staircase labeled 1 is 8. There is a pair of these staircases, so T(15,1) = 2*8 = 16, since the symmetric representation of sigma(15) is also the base of the three dimensional version of the ziggurat . %e A347529 The number of steps in the double-staircase labeled 3 is equal to 7, so T(15,2) = 7. %e A347529 The number of steps in the double-staircase labeled 5 is equal to 1, so T(15,3) = 1. %e A347529 Therefore the 15th row of triangle is [16, 7, 1]. %e A347529 The top view of the 3D-Ziggurat of 15 and the symmetric representation of sigma(15) with subparts look like this: %e A347529 _ _ %e A347529 |_| | | %e A347529 |_| | | %e A347529 |_| | | %e A347529 |_| | | %e A347529 |_| | | %e A347529 |_| | | %e A347529 |_| | | %e A347529 _ _ _|_| _ _ _|_| %e A347529 _ _|_| 36 _ _| | 8 %e A347529 |_|_|_| | _ _| %e A347529 _|_|_| _| |_| %e A347529 |_|_| 1 |_ _| 1 %e A347529 | 34 | 7 %e A347529 _ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _| %e A347529 |_|_|_|_|_|_|_|_| |_ _ _ _ _ _ _ _| %e A347529 36 8 %e A347529 . %e A347529 Top view of the 3D-Ziggurat. The symmetric representation of %e A347529 The ziggurat is formed by 3 of sigma(15) is formed by 3 parts. %e A347529 polycubes with 107 cubes It has 4 subparts with 24 cells in %e A347529 in total. It has 4 staircases total. It is the base of the ziggurat. %e A347529 with 24 steps in total. %e A347529 . %Y A347529 Another (and more regular) version of A279388. %Y A347529 Row sums give A000203. %Y A347529 Row n has length A351846(n). %Y A347529 Cf. A347263 (analog for the ziggurat diagram). %Y A347529 Cf. A000040, A000079, A000384, A000396, A082647, A196020, A235791, A236104, A237591, A237270, A237271, A237593, A279387, A262626, A335616, A346875, A347186. %K A347529 nonn,tabf %O A347529 1,2 %A A347529 _Omar E. Pol_, Sep 05 2021