

A237591


Triangle read by rows: T(n,k) = A235791(n,k)  A235791(n,k+1), assuming that the virtual right border of triangle A235791 is A000004.


241



1, 2, 2, 1, 3, 1, 3, 2, 4, 1, 1, 4, 2, 1, 5, 2, 1, 5, 2, 2, 6, 2, 1, 1, 6, 3, 1, 1, 7, 2, 2, 1, 7, 3, 2, 1, 8, 3, 1, 2, 8, 3, 2, 1, 1, 9, 3, 2, 1, 1, 9, 4, 2, 1, 1, 10, 3, 2, 2, 1, 10, 4, 2, 2, 1, 11, 4, 2, 1, 2, 11, 4, 3, 1, 1, 1, 12, 4, 2, 2, 1, 1, 12, 5, 2, 2, 1, 1, 13, 4, 3, 2, 1, 1, 13, 5, 3, 1, 2, 1, 14, 5, 2, 2, 2, 1
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OFFSET

1,2


COMMENTS

T(n,k) is also the length of the kth segment in a zigzag path on the first quadrant of the square grid, connecting the point (n, 0) with the point (m, m), starting with a segment in vertical direction, where m <= n.
Conjecture: the area of the polygon defined by the xaxis, this zigzag path and the diagonal [(0, 0), (m, m)], is equal to A024916(n)/2, one half of the sum of all divisors of all positive integers <= n. Therefore the reflected polygon, which is adjacent to the yaxis, with the zigzag path connecting the point (0, n) with the point (m, m), has the same property. And so on for each octant in the four quadrants.
For the representation of A024916 and A000203 we use two octants, for example: the first octant and the second octant, or the 6th octant and the 7th octant, etc., see A237593.
At least up to n = 128, two zigzag paths never cross (checked by hand).
The finite sequence formed by the nth row of triangle together with its mirror row gives the nth row of triangle A237593.
The connection between A196020 and A237271 is as follows: A196020 > A236104 > A235791 > this sequence > A237593 > A239660 > A237270 > A237271.
Comments from Franklin T. AdamsWatters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014. (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
The row sum is A235791(n,1)  A235791(n,floor((sqrt(8n+1)1)/2)+1) = n  0.
Mathematica function has been written to check the conjecture as well as noncrossing zigzag paths (Dyck paths rotated by 90 degrees) up through n=30000 (same applies to A237593). (End)
The nth zigzag path ending at the point (m, m), where m = A240542(n).  Omar E. Pol, Apr 16 2014
From Omar E. Pol, Aug 23 2015: (Start)
n is an odd prime if and only if T(n,2) = 1 + T(n1,2) and T(n,k) = T(n1,k) for the rest of the values of k.
The elements of the nth row of triangle together with the elements of the nth row of triangle A261350 give the nth row of triangle A237593.
T(n,k) is also the area (or the number of cells) of the kth vertical side at the nth level (starting from the top) in the left hand part of the front view of the stepped pyramid described in A245092, see Example section.
(End)
From Omar E. Pol, Nov 19 2015: (Start)
T(n,k) is also the number of cells between the kth and the (k+1)st line segments (from left to right) in the nth row of the diagram as shown in Example section.
Note that the number of horizontal line segments in the nth row of the diagram equals A001227(n), the number of odd divisors of n. (End)
Conjecture: the values f(n,k) in the nth row of the triangle are either 1 or 2 for all k with ceiling((sqrt(4*n+1)1)/2) <= k <= floor((sqrt(8*n+1)1)/2) = r(n), the length of the nth row, though the lower bound need not be minimal; tested through 2500000. See also A285356.  Hartmut F. W. Hoft, Apr 17 2017
Conjecture: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts.  Omar E. Pol, Apr 30 2017
From Omar E. Pol, Aug 31 2021: (Start)
It appears that T(n,2)/T(n,1) converges to 1/3.
It appears that T(n,3)/T(n,2) converges to 1/2.
It appears that T(n,4)/T(n,3) converges to 3/5.
It appears that T(n,5)/T(n,4) converges to 2/3. (End)
In other words: T(n,k) is the length of the kth line segment of the largest Dyck path of the symmetric representation of sigma(n).  Omar E. Pol, Sep 08 2021


LINKS

G. C. Greubel, Table of n, a(n) for the first 150 rows


FORMULA

T(n,k) = ceiling((n+1)/k  (k+1)/2)  ceiling((n+1)/(k+1)  (k+2)/2), for 1 <= n and 1 <= k <= floor((sqrt(8n+1)1)/2).  Hartmut F. W. Hoft, Apr 07 2014


EXAMPLE

Triangle begins:
1;
2;
2, 1;
3, 1;
3, 2;
4, 1, 1;
4, 2, 1;
5, 2, 1;
5, 2, 2;
6, 2, 1, 1;
6, 3, 1, 1;
7, 2, 2, 1;
7, 3, 2, 1;
8, 3, 1, 2;
8, 3, 2, 1, 1;
9, 3, 2, 1, 1;
9, 4, 2, 1, 1;
10, 3, 2, 2, 1;
10, 4, 2, 2, 1;
11, 4, 2, 1, 2;
11, 4, 3, 1, 1, 1;
12, 4, 2, 2, 1, 1;
12, 5, 2, 2, 1, 1;
13, 4, 3, 2, 1, 1;
13, 5, 3, 1, 2, 1;
14, 5, 2, 2, 2, 1;
14, 5, 3, 2, 1, 2;
15, 5, 3, 2, 1, 1, 1;
...
For n = 10 the 10th row of triangle A235791 is [10, 4, 2, 1] so row 10 is [6, 2, 1, 1].
From Omar E. Pol, Aug 23 2015: (Start)
Illustration of initial terms:
Row _
1 _1
2 _2 _
3 _2 1
4 _3 _1
5 _3 2 _
6 _4 _11
7 _4 2 1
8 _5 _2 _1
9 _5 2 2 _
10 _6 _2 11
11 _6 3 _11
12 _7 _2 2 1
13 _7 3 2 _1
14 _8 _3 _12 _
15 _8 3 2 11
16 _9 _3 2 11
17 _9 4 _2 _11
18 _10 _3 2 2 1
19 _10 4 2 2 _1
20 _11 _4 _2 12 _
21 _11 4 3 _111
22 _12 _4 2 2 11
23 _12 5 _2 2 11
24 _13 _4 3 2 _11
25 _13 5 3 _12 1
26 _14 _5 _2 2 2 _1
27 _14 5 3 2 12 _
28 15 5 3 2 111
...
Also the diagram represents the left part of the front view of the pyramid described in A245092. For the other half front view see A261350. For more information about the pyramid and the symmetric representation of sigma see A237593. (End)
From Omar E. Pol, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below: _
 
 
 
 
 
_ _ _ 
_ _ _
_ 
 _
 _ _1
_ _ _ _ _ _  2
_ _ _ _ _ _ _2
7
.
The lengths of the successive line segments from the first vertex to the central vertex of the largest Dyck path are [7, 2, 2, 1] respectively, the same as the 12th row of triangle. (End)


MATHEMATICA

row[n_]:= Floor[(Sqrt[8*n+1] 1)/2]; f[n_, k_]:= Ceiling[(n+1)/k(k+1)/2]  Ceiling[(n+1)/(k+1)(k+2)/2];
Table[f[n, k], {n, 1, 50}, {k, 1, row[n]}]//Flatten
(* Hartmut F. W. Hoft, Apr 08 2014 *)


PROG

(PARI) row235791(n) = vector((sqrtint(8*n+1)1)\2, i, 1+(n(i*(i+1)/2))\i);
row(n) = {my(orow = concat(row235791(n), 0)); vector(#orow 1, i, orow[i]  orow[i+1]); } \\ Michel Marcus, Mar 27 2014
(Python)
from sympy import sqrt
import math
def T(n, k): return int(math.ceil((n + 1)/k  (k + 1)/2))  int(math.ceil((n + 1)/(k + 1)  (k + 2)/2))
for n in range(1, 29): print([T(n, k) for k in range(1, int((sqrt(8*n + 1)  1)/2) + 1)]) # Indranil Ghosh, Apr 30 2017


CROSSREFS

Row n has length A003056(n) hence column k starts in row A000217(k).
Row sums give A000027.
Column 1 is A008619, n >= 1.
Cf. A000203, A001227, A024916, A196020, A235791, A236104, A237048, A237270, A237271, A237593, A239660, A239931A239934, A240542, A244580, A245092, A249351, A259176, A259177, A261350, A261699, A262626, A285356, A286000, A286001.
Sequence in context: A167413 A340985 A259176 * A277730 A174167 A159876
Adjacent sequences: A237588 A237589 A237590 * A237592 A237593 A237594


KEYWORD

nonn,tabf


AUTHOR

Omar E. Pol, Feb 22 2014


EXTENSIONS

3 more rows added by Omar E. Pol, Aug 23 2015


STATUS

approved



