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%I #94 Mar 05 2022 00:34:35
%S 1,3,2,2,7,3,3,11,1,4,4,15,5,3,5,9,9,6,6,23,5,7,7,12,12,8,7,1,8,31,9,
%T 9,35,2,2,10,10,39,3,11,5,5,11,18,18,12,12,47,13,13,5,13,21,21,14,6,6,
%U 14,55,1,15,15,59,3,7,3,16,16,63,17,7,7,17,27,27,18,9,3,18,71,10,10,19,19,30,30
%N Irregular triangle read by rows in which row n lists the subparts of the symmetric representation of sigma(n), ordered by order of appearance in the structure, from left to right.
%C The terms in the n-th row are the same as the terms in the n-th row of triangle A279391, but in some rows the terms appear in distinct order.
%C First differs from A279391 at a(28) = T(15,3).
%C Also nonzero terms of A296508. - _Omar E. Pol_, Feb 11 2018
%H Hartmut F. W. Hoft, <a href="/A280851/b280851.txt">Table of n, a(n) for n = 1..987</a>
%H Hartmut F. W. Hoft, <a href="/A280851/a280851.pdf">A proof that the symmetric representation of sigma equals sigma</a>
%H Hartmut F. W. Hoft, <a href="/A280851/a280851_1.pdf">Numbers having a subpart of size 1 in the symmetric representation of sigma are the hexagonal numbers</a>, A proof of _Omar E. Pol_'s conjecture dated Aug 28 2021 in A000384.
%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
%e Triangle begins (rows 1..16):
%e 1;
%e 3;
%e 2, 2;
%e 7;
%e 3, 3;
%e 11, 1;
%e 4, 4;
%e 15;
%e 5, 3, 5;
%e 9, 9;
%e 6, 6;
%e 23, 5;
%e 7, 7;
%e 12, 12;
%e 8, 7, 1, 8;
%e 31;
%e ...
%e For n = 12 we have that the 11th row of triangle A237593 is [6, 3, 1, 1, 1, 1, 3, 6] and the 12th row of the same triangle is [7, 2, 2, 1, 1, 2, 2, 7], so the diagram of the symmetric representation of sigma(12) = 28 is constructed as shown below in Figure 1:
%e . _ _
%e . | | | |
%e . | | | |
%e . | | | |
%e . | | | |
%e . | | | |
%e . _ _ _| | _ _ _| |
%e . _| _ _| _| _ _ _|
%e . _| | _| _| |
%e . | _| | _| _|
%e . | _ _| | |_ _|
%e . _ _ _ _ _ _| | 28 _ _ _ _ _ _| | 5
%e . |_ _ _ _ _ _ _| |_ _ _ _ _ _ _|
%e . 23
%e .
%e . Figure 1. The symmetric Figure 2. After the dissection
%e . representation of sigma(12) of the symmetric representation
%e . has only one part which of sigma(12) into layers of
%e . contains 28 cells, so width 1 we can see two subparts
%e . the 12th row of the that contain 23 and 5 cells
%e . triangle A237270 is [28]. respectively, so the 12th row of
%e . this triangle is [23, 5].
%e .
%e For n = 15 we have that the 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8], so the diagram of the symmetric representation of sigma(15) = 24 is constructed as shown below in Figure 3:
%e . _ _
%e . | | | |
%e . | | | |
%e . | | | |
%e . | | | |
%e . | | | |
%e . | | | |
%e . | | | |
%e . _ _ _|_| _ _ _|_|
%e . _ _| | 8 _ _| | 8
%e . | _| | _ _|
%e . _| _| _| |_|
%e . |_ _| 8 |_ _| 1
%e . | | 7
%e . _ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _|
%e . |_ _ _ _ _ _ _ _| |_ _ _ _ _ _ _ _|
%e . 8 8
%e .
%e . Figure 3. The symmetric Figure 4. After the dissection
%e . representation of sigma(15) of the symmetric representation
%e . has three parts of size 8 of sigma(15) into layers of
%e . because every part contains width 1 we can see four "subparts".
%e . 8 cells, so the 15th row of The first layer has three subparts:
%e . triangle A237270 is [8, 8, 8]. [8, 7, 8]. The second layer has
%e . only one subpart of size 1. The
%e . 15th row of this triangle is
%e . [8, 7, 1, 8].
%e .
%e From _Hartmut F. W. Hoft_, Jan 31 2018: (Start)
%e The subparts of 36 whose symmetric representation of sigma has maximum width 2 are 71, 10, and 10.
%e The (size, width level) pairs of the six subparts of the symmetric representation of sigma(63) which consists of five parts are (32,1), (12,1), (11,1), (5,2), (12,1), and (32,1).
%e The subparts of perfect number 496 are 991, the length of its entire Dyck path, and 1 at the diagonal.
%e Number 10080, the smallest number whose symmetric representation of sigma has maximum width 10 (see A250070), has 12 subparts; its (size, width level) pairs are (20159,1), (6717,2), (4027,3), (2873,4), (2231,5), (1329,6), (939,7), (541,8), (403,9), (3,10), (87,10), and (3,10). The size of the first subpart is the length of the entire Dyck path so that the symmetric representation consists of a single part. The first subpart at the 10th level occurs at coordinates (6926,7055) ... (6929,7055). (End)
%e From _Omar E. Pol_, Dec 26 2020: (Start)
%e Also consider the infinite double-staircases diagram defined in A335616 (see the theorem).
%e For n = 15 the diagram with first 15 levels looks like this:
%e .
%e Level "Double-staircases" diagram
%e . _
%e 1 _|1|_
%e 2 _|1 _ 1|_
%e 3 _|1 |1| 1|_
%e 4 _|1 _| |_ 1|_
%e 5 _|1 |1 _ 1| 1|_
%e 6 _|1 _| |1| |_ 1|_
%e 7 _|1 |1 | | 1| 1|_
%e 8 _|1 _| _| |_ |_ 1|_
%e 9 _|1 |1 |1 _ 1| 1| 1|_
%e 10 _|1 _| | |1| | |_ 1|_
%e 11 _|1 |1 _| | | |_ 1| 1|_
%e 12 _|1 _| |1 | | 1| |_ 1|_
%e 13 _|1 |1 | _| |_ | 1| 1|_
%e 14 _|1 _| _| |1 _ 1| |_ |_ 1|_
%e 15 |1 |1 |1 | |1| | 1| 1| 1|
%e .
%e Starting from A196020 and after the algorithm described in A280850 and A296508 applied to the above diagram we have a new diagram as shown below:
%e .
%e Level "Ziggurat" diagram
%e . _
%e 6 |1|
%e 7 _ | | _
%e 8 _|1| _| |_ |1|_
%e 9 _|1 | |1 1| | 1|_
%e 10 _|1 | | | | 1|_
%e 11 _|1 | _| |_ | 1|_
%e 12 _|1 | |1 1| | 1|_
%e 13 _|1 | | | | 1|_
%e 14 _|1 | _| _ |_ | 1|_
%e 15 |1 | |1 |1| 1| | 1|
%e .
%e The 15th row
%e of A249351 : [1,1,1,1,1,1,1,1,0,0,0,1,1,1,2,1,1,1,0,0,0,1,1,1,1,1,1,1,1]
%e The 15th row
%e of A237270: [ 8, 8, 8 ]
%e The 15th row
%e of A296508: [ 8, 7, 1, 0, 8 ]
%e The 15th row
%e of triangle [ 8, 7, 1, 8 ]
%e .
%e More generally, for n >= 1, it appears there is the same correspondence between the original diagram of the symmetric representation of sigma(n) and the "Ziggurat" diagram of n.
%e For the definition of subparts see A279387 and also A296508. (End)
%t row[n_] := Floor[(Sqrt[8n+1]-1)/2]
%t f[n_] := Map[Ceiling[(n+1)/#-(#+1)/2] - Ceiling[(n+1)/(#+1)-(#+2)/2]&, Range[row[n]]]
%t a237593[n_] := Module[{a=f[n]}, Join[a, Reverse[a]]]
%t g[n_] := Map[If[Mod[n - #*(#+1)/2, #]==0, (-1)^(#+1), 0]&, Range[row[n]]]
%t a262045[n_] := Module[{a=Accumulate[g[n]]}, Join[a, Reverse[a]]]
%t findStart[list_] := Module[{i=1}, While[list[[i]]==0, i++]; i]
%t a280851[n_] := Module[{lenL=a237593[n], widL=a262045[n], r=row[n], subs={}, acc, start, i}, While[!AllTrue[widL, #==0&], start=findStart[widL]; acc=lenL[[start]]; widL[[start]]-=1; i=start+1; While[i<=2*r && acc!=0, If[widL[[i]]==0, If[start<=r<i, AppendTo[subs, acc-1], AppendTo[subs, acc]]; acc=0, acc+=lenL[[i]]; widL[[i]]-=1; i++]]; If[i>2*r && acc!=0, If[start<=r<i, AppendTo[subs, acc-1], AppendTo[subs, acc]]; acc=0]]; subs]
%t Flatten[Map[a280851, Range[36]]] (* data *)
%t TableForm[Map[{#, a280851[#]}&, Range[36]], TableDepth->2] (* triangle *) (* _Hartmut F. W. Hoft_, Jan 31 2018 *)
%Y Row sums give A000203.
%Y The length of row n equals A001227(n).
%Y Hence, if n is odd the length of row n equals A000005(n).
%Y For the definition of "subparts" see A279387.
%Y For the triangle of sums of subparts see A279388.
%Y Cf. A000384, A001227, A196020, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A239657, A240542, A244050, A245092, A249351, A250068, A250070, A261699, A262626, A279391, A280850, A296508, A335616, A346875.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Jan 09 2017
%E Name clarified by _Hartmut F. W. Hoft_ and _Omar E. Pol_, Jan 31 2018
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