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Row n is a palindromic composition of 2*n.

T(n,k) is also the length of the k-th segment in a Dyck path on the first quadrant of the square grid, connecting the x-axis with the y-axis, from (n, 0) to (0, n), starting with a segment in vertical direction, see example.

Conjecture 1: the area under the n-th Dyck path equals A024916(n), the sum of all divisors of all positive integers <= n.

If the conjecture is true then the n-th Dyck path represents the boundary segments after the alternating sum of the elements of the n-th row of A236104.

Conjecture 2: two adjacent Dyck paths never cross (checked by hand up to n = 128), hence the total area between the n-th Dyck path and the (n-1)-st Dyck path is equal to sigma(n) = A000203(n), the sum of divisors of n.

The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> this sequence --> A239660 --> A237270 --> A237271.

PARI scripts area(n) and chkcross(n) have been written to check the 2 properties and have been run up to n=10000. - Michel Marcus, Mar 27 2014

Mathematica functions have been written that verified the 2 properties through n=30000. - Hartmut F. W. Hoft, Apr 07 2014

Comments from Franklin T. Adams-Watters 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 n-1; and the number of squares added is sigma(n).

Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)

The symmetric representation of sigma, so defined, is row n of A237270. - Peter Munn, Jan 06 2025

It appears that, for the n-th set, the number of cells lying on the first diagonal is equal to A067742(n), the number of middle divisors of n. - Michel Marcus, Jun 21 2014

Checked Michel Marcus's conjecture with two Mathematica functions up to n=100000, for more information see A240542. - Hartmut F. W. Hoft, Jul 17 2014

A003056(n) is also the number of peaks of the Dyck path related to the n-th row of triangle. - Omar E. Pol, Nov 03 2015

The number of peaks of the Dyck path associated to the row A000396(n) of this triangle equals the n-th Mersenne prime A000668(n), hence Mersenne primes are visible in two ways at the pyramid described in A245092. - Omar E. Pol, Dec 19 2016

The limit as n approaches infinity (area under the Dyck path described in the n-th row of triangle divided by n^2) equals Pi^2/12 = zeta(2)/2. (Cf. A072691.) - Omar E. Pol, Dec 18 2021

The connection between the isosceles triangle and the stepped pyramid is due to the fact that this object can also be interpreted as a pop-up card. - Omar E. Pol, Nov 09 2022

T(n,k) is the number of cells in the k-th region of the n-th set of regions in a diagram of the symmetry of sigma(n), see example.

Row n is a palindromic composition of sigma(n).

Row sums give A000203.

Row n has length A237271(n).

In the row 2n-1 of triangle both the first term and the last term are equal to n.

If n is an odd prime then row n is [m, m], where m = (1 + n)/2.

The connection with A196020 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> this sequence.

For the boundary segments in an octant see A237591.

For the boundary segments in a quadrant see A237593.

For the boundary segments in the spiral see also A239660.

For the parts in every quadrant of the spiral see A239931, A239932, A239933, A239934.

We can find the spiral on the terraces of the stepped pyramid described in A244050. - Omar E. Pol, Dec 07 2016

T(n,k) is also the area of the k-th terrace, from left to right, at the n-th level, starting from the top, of the stepped pyramid described in A245092 (see Links section). - Omar E. Pol, Aug 14 2018

The original name was: 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.

T(n,k) is also the length of the k-th segment in a zig-zag 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 x-axis, this zig-zag 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 y-axis, with the zig-zag 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 zig-zag paths never cross (checked by hand).

The finite sequence formed by the n-th row of triangle together with its mirror row gives the n-th 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. Adams-Watters 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 n-1; and the number of squares added is sigma(n).

Finally, look at the connected areas enclosed by n but not by n-1; 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 non-crossing zig-zag paths (Dyck paths rotated by 90 degrees) up through n=30000 (same applies to A237593). (End)

The n-th zig-zag 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(n-1,2) and T(n,k) = T(n-1,k) for the rest of the values of k.

The elements of the n-th row of triangle together with the elements of the n-th row of triangle A261350 give the n-th row of triangle A237593.

T(n,k) is also the area (or the number of cells) of the k-th vertical side at the n-th 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 k-th and the (k+1)st line segments (from left to right) in the n-th row of the diagram as shown in Example section.

Note that the number of horizontal line segments in the n-th row of the diagram equals A001227(n), the number of odd divisors of n. (End)

Conjecture: the values f(n,k) in the n-th 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 n-th 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 k-th line segment of the largest Dyck path of the symmetric representation of sigma(n). - Omar E. Pol, Sep 08 2021

Consider an irregular stepped pyramid with n steps. The base of the pyramid is equal to the symmetric representation of A024916(n), the sum of all divisors of all positive integers <= n. Two of the faces of the pyramid are the same as the representation of the n-th triangular numbers as a staircase. The total area of the pyramid is equal to 2*A024916(n) + A046092(n). The volume is equal to A175254(n). By definition a(2n-1) is A000203(n), the sum of divisors of n. Starting from the top a(2n-1) is also the total area of the horizontal part of the n-th step of the pyramid. By definition, a(2n) = A005843(n) = 2n. Starting from the top, a(2n) is also the total area of the irregular vertical part of the n-th step of the pyramid.

On the other hand the sequence also has a symmetric representation in two dimensions, see Example.

From Omar E. Pol, Dec 31 2016: (Start)

We can find the pyramid after the following sequences: A196020 --> A236104 --> A235791 --> A237591 --> A237593.

The structure of this infinite pyramid arises after the 90-degree-zig-zag folding of the diagram of the isosceles triangle A237593 (see the links).

The terraces at the m-th level of the pyramid are also the parts of the symmetric representation of sigma(m), m >= 1, hence the sum of the areas of the terraces at the m-th level equals A000203(m).

Note that the stepped pyramid is also one of the 3D-quadrants of the stepped pyramid described in A244050.

For more information about the pyramid see A237593 and all its related sequences. (End)

a(n) is also the volume of a special stepped pyramid with n levels related to the symmetric representation of sigma. Note that starting at the top of the pyramid, the total area of the horizontal regions at the n-th level is equal to A239050(n), and the total area of the vertical regions at the n-th level is equal to 8*n.

From Omar E. Pol, Sep 19 2015: (Start)

Also, consider that the area of the central square in the top of the pyramid is equal to 1, so the total area of the horizontal regions at the n-th level starting from the top is equal to sigma(n) = A000203(n), and the total area of the vertical regions at the n-th level is equal to 2*n.

Also note that this stepped pyramid can be constructed with four copies of the stepped pyramid described in A245092 back-to-back (one copy in every quadrant). (End)

From Omar E. Pol, Jan 20 2021: (Start)

Convolution of A000203 and the nonzero terms of A008586.

Convolution of A074400 and the nonzero terms of A005843.

Convolution of A340793 and the nonzero terms of A046092.

Convolution of A239050 and A000027.

(End)

For the construction of this sequence also we can start from A235791.

This sequence can be interpreted as an infinite Dyck path: UDUDUUDD...

Also we use this sequence for the construction of a spiral in which the arms in the quadrants give the symmetric representation of sigma, see example.

We can find the spiral (mentioned above) on the terraces of the stepped pyramid described in A244050. - Omar E. Pol, Dec 07 2016

The spiral has the property that the sum of the parts in the quadrants 1 and 3, divided by the sum of the parts in the quadrants 2 and 4, converges to 3/5. - Omar E. Pol, Jun 10 2019

Row n is a palindromic composition of sigma(4n).

Row n is also the row 4n of A237270.

Row n has length A237271(4n).

Row sums give A193553.

First differs from A193553 at a(11).

Also row n lists the parts of the symmetric representation of sigma in the n-th arm of the fourth quadrant of the spiral described in A239660, see example.

For the parts of the symmetric representation of sigma(4n-3), see A239931.

For the parts of the symmetric representation of sigma(4n-2), see A239932.

For the parts of the symmetric representation of sigma(4n-1), see A239933.

We can find the spiral (mentioned above) on the terraces of the pyramid described in A244050. - Omar E. Pol, Dec 06 2016

Conjecture: row n is formed by the odd-indexed terms of the n-th row of triangle A280850 together with the even-indexed terms of the same row but listed in reverse order. Examples: the 15th row of A280850 is [8, 8, 7, 0, 1] so the 15th row of this triangle is [8, 7, 1, 0, 8]. The 75th row of A280850 is [38, 38, 21, 0, 3, 3, 0, 0, 0, 21, 0] so the 75h row of this triangle is [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38].

For the definition of "subparts" see A279387.

For more information about the mentioned Dyck paths see A237593.

T(n,k) could be called the "charge" of the k-th peak of the largest Dyck path of the symmetric representation of sigma(n).

The number of zeros in row n is A238005(n). - Omar E. Pol, Sep 11 2021

All odd primes are in the sequence because the parts of the symmetric representation of sigma(prime(i)) are [m, m], where m = (1 + prime(i))/2, for i >= 2.

There are no odd composite numbers in this sequence.

First differs from A173708 at a(13).

Since sigma(p*q) >= 1 + p + q + p*q for odd p and q, the symmetric representation of sigma(p*q) has more parts than the two extremal ones of size (p*q + 1)/2; therefore, the above comments are true. - Hartmut F. W. Hoft, Jul 16 2014

From Hartmut F. W. Hoft, Sep 16 2015: (Start)

The following two statements are equivalent:

(1) The symmetric representation of sigma(n) has two parts, and

(2) n = q * p where q is in A174973, p is prime, and 2 * q < p.

For a proof see the link and also the link in A071561.

This characterization allows for much faster computation of numbers in the sequence - function a239929F[] in the Mathematica section - than computations based on Dyck paths. The function a239929Stalk[] gives rise to the associated irregular triangle whose columns are indexed by A174973 and whose rows are indexed by A065091, the odd primes. (End)

From Hartmut F. W. Hoft, Dec 06 2016: (Start)

For the respective columns of the irregular triangle with fixed m: k = 2^m * p, m >= 1, 2^(m+1) < p and p prime:

(a) each number k is representable as the sum of 2^(m+1) but no fewer consecutive positive integers [since 2^(m+1) < p].

(b) each number k has 2^m as largest divisor <= sqrt(k) [since 2^m < sqrt(k) < p].

(c) each number k is of the form 2^m * p with p prime [by definition].

m = 1: (a) A100484 even semiprimes (except 4 and 6)

(b) A161344 (except 4, 6 and 8)

(c) A001747 (except 2, 4 and 6)

m = 2: (a) A270298

(b) A161424 (except 16, 20, 24, 28 and 32)

(c) A001749 (except 8, 12, 20 and 28)

m = 3: (a) A270301

(b) A162528 (except 64, 72, 80, 88, 96, 104, 112 and 128)

(c) sequence not in OEIS

b(i,j) = A174973(j) * {1,5) mod 6 * A174973(j), for all i,j >= 1; see A091999 for j=2. (End)

Conjecture 1: where records occur in A237271. - Omar E. Pol, Dec 27 2016

For more information about the symmetric representation of sigma see A237270, A237593.

This sequence of (first occurrence of) parts appears to be strictly increasing in contrast to sequence A250070 of (first occurrence of) maximum widths. - Hartmut F. W. Hoft, Dec 09 2014

Conjecture 2: all terms are odd numbers. - Omar E. Pol, Oct 14 2018

Proof of Conjecture 2: Let n = 2^m * q with m>0 and q odd; then the 1's in even positions of row n in the triangle of A237048 are at positions 2^(m+1) * d <= row(n) where d divides q. For n/2 the even positions of 1's occur at the smaller values 2^m * d <= row(n/2), thus either keeping or reducing widths (A249223) of parts in the symmetric representation of sigma for n/2 inherited from row n. Therefore the number of parts for n is at most as large as for n/2, i.e., all numbers in this sequence are odd. - Hartmut F. W. Hoft, Sep 22 2021

Observation: at least for n = 1..21 we have that 2*a(n) < a(n+1). - Omar E. Pol, Sep 22 2021

From Omar E. Pol, Jul 28 2025: (Start)

Conjecture 3: a(n) is the smallest number k having n 2-dense sublists of divisors of k.

The 2-dense sublists of divisors of k are the maximal sublists whose terms increase by a factor of at most 2.

In a sublist of divisors of k the terms are in increasing order and two adjacent terms are the same two adjacent terms in the list of divisors of k.

An example of the conjecture 3 for n = 1..5 is as shown below:

----------------------------------------------------

| | List of divisors of k | | |

| k | [with sublists in brackets] | n | a(n) |

----------------------------------------------------

| 1 | [1]; | 1 | 1 |

| 3 | [1], [3]; | 2 | 3 |

| 9 | [1], [3], [9]; | 3 | 9 |

| 21 | [1], [3], [7], [21]; | 4 | 21 |

| 63 | [1], [3], [7, 9], [21], [63]; | 5 | 63 |

(End)

Conjecture 4: a(n) is the smallest number k having n divisors p of k such that p is greater than twice the adjacent previous divisor of k. - Omar E. Pol, Aug 05 2025