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Number n is in this sequence exactly when two parts of the symmetric representation of sigma(n) meet at the diagonal.

Proof: If n = k*(2*k+1) is in this sequence then the length of row n in A240542 is 2*k and that of row n-1 is 2*k-1, i.e., the last leg of the Dyck path for n down to the diagonal is vertical and that for n-1 is horizontal to the same point on the diagonal. Therefore, one part of the symmetric representation of sigma(n) ends at the diagonal and so does its symmetric copy. Conversely, if two parts meet at the diagonal then the number of legs in the Dyck path to the diagonal for n, i.e., the length of row n in A240542, is one larger than that for n-1 and must be even, i.e., n has the form n = k*(2*k+1).

A156592 is a subsequence since for every number of the form n = p*(2*p+1) where both p and 2*p+1 are primes A240542(n) = A240542(n-1). For a proof let T(n,k) = ceiling((n+1)/k - (k+1)/2) for 1 <= k <= floor((sqrt(8*n+1) - 1)/2) = 2*p, see A235791; then T(n,k) = T(n-1,k) + 1 for k = 1, 2, p, 2*p, and T(n,k) = T(n-1,k) for all other k. Therefore, the two alternating sums defining A240542(n) and A240542(n-1) are equal, i.e., their Dyck paths meet at the diagonal.

Except for missing 10 the intersection of this sequence and A298855 equals A156592. Sequence A262259 is a subsequence of this sequence.

The five known members of A191363 belong to this sequence, and since their symmetric representation consists of two parts of width one (the respective rows of triangle A237048 have the form 1 0 ... 0 1) they also belong to A262259.

Subsequence of A014105. - Omar E. Pol, Jan 31 2018

Second hexagonal numbers without middle divisors. - Omar E. Pol, Mar 10 2023

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

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

The alternating sum of the squares of the elements of the n-th row equals the sum of all divisors of all positive integers <= n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*(T(n,k))^2 = A024916(n).

Row n has length A003056(n) hence the first element of column k is in row A000217(k).

For more information see A236104.

The sum of row n gives A060831(n), the sum of the number of odd divisors of all positive integers <= n. - Omar E. Pol, Mar 01 2014. [An equivalent assertion is that the sum of row n of A237048 is the number of odd divisors of n, and this was proved by Hartmut F. W. Hoft in a comment in A237048. - N. J. A. Sloane, Dec 07 2020]

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)

Mathematica function has been written to check the first property up to n = 20000.

T(n,(sqrt(8n+1)-1)/2+1) = 0 for all n >= 1, which is useful for formulas for A237591 and A237593. (End)

Alternating row sums give A240542. - Omar E. Pol, Apr 16 2014

Conjecture: T(n,k) is also the total number of partitions of all positive integers <= n into exactly k consecutive parts, i.e., the partial column sum of A285898, or in accordance with the triangles of the same family: the partial column sum of A237048. - Omar E. Pol, Apr 28 2017, Nov 24 2020

The above conjecture is true. The proof will be added soon (it uses the generating function for the columns). - N. J. A. Sloane, Nov 24 2020

T(n,k) is also the total length of all line segments between the k-th vertex and the central vertex of the largest Dyck path of the symmetric representation of sigma(n). In other words: T(n,k) is the sum of the last (A003056(n)-k+1) terms of the n-th row of A237591. - Omar E. Pol, Sep 07 2021

T(n,k) is also the Manhattan distance between the k-th vertex and the central vertex of the Dyck path described in the n-th row of the triangle A237593. - Omar E. Pol, Jan 11 2023

Comment from N. J. A. Sloane, Jan 03 2021: (Start)

Theorem 1: (i) a(n) = (number of odd divisors of n <= sqrt(2*n)) - (number of odd divisors of n > sqrt(2*n)).

(ii) Let r(n) = A003056(n). Then a(n) = (number of odd divisors of n <= r(n)) - (number of odd divisors of n > r(n)).

(iii) a(n) = Sum_{k=1..r(n)} (-1)^(k+1)*A237048(n,k).

(iv) a(n) is odd iff n is a square or twice a square (cf. A053866). Indices of odd terms give A028982. Indices of even terms give A028983.

The proofs are straightforward. These results were conjectured by Omar E. Pol in 2017. (End)

Theorem 2: a(n) is equal to the difference between the number of partitions of n into an odd number of consecutive parts and the number of partitions of n into an even number of consecutive parts. [Chapman et al., 2001; Hirschhorn and Hirschhorn, 2005]. - Omar E. Pol, Feb 06 2017

From Omar E. Pol, Feb 06 2017: (Start)

Conjecture 1: This is the central column of the isosceles triangle of A249351.

Conjecture 2: a(n) is also the width of the terrace at the n-th level in the main diagonal of the pyramid described in A245092.

Conjecture 3: a(n) is also the number of central subparts of the symmetric representation of sigma(n). For more information see A279387.

Conjectures 2 and 3 were proposed after Michel Marcus's conjecture in A237593. (End)

Conjectures 1, 2, and 3 are all true. - N. J. A. Sloane, Feb 11 2021

Here T(n,k) is defined to be the "k-th width" of the symmetric representation of sigma(n), with n>=1 and 1<=k<=2n-1. Explanation: consider the diagram of the symmetric representation of sigma(n) described in A236104, A237593 and other related sequences. Imagine that the diagram for sigma(n) contains 2n-1 equidistant segments which are parallel to the main diagonal [(0,0),(n,n)] of the quadrant. The segments are located on the diagonal of the cells. The distance between two parallel segment is equal to sqrt(2)/2. T(n,k) is the length of the k-th segment divided by sqrt(2). Note that the triangle contains nonnegative terms because for some n the value of some widths is equal to zero. For an illustration of some widths see Hartmut F. W. Hoft's contribution in the Links section of A237270.

Row n has length 2*n-1.

Row sums give A000203.

If n is a power of 2 then all terms of row n are 1's.

If n is an even perfect number then all terms of row n are 1's except the middle term which is 2.

If n is an odd prime then row n lists (n+1)/2 1's, n-2 zeros, (n+1)/2 1's.

The number of blocks of positive terms in row n gives A237271(n).

The sum of the k-th block of positive terms in row n gives A237270(n,k).

It appears that the middle diagonal is also A067742 (which was conjectured by Michel Marcus in the entry A237593 and checked with two Mathematica functions up to n = 100000 by Hartmut F. W. Hoft).

It appears that the trapezoidal numbers (A165513) are also the numbers k > 1 with the property that some of the noncentral widths of the symmetric representation of sigma(k) are not equal to 1. - Omar E. Pol, Mar 04 2023

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

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.

First differs from A279391 at a(28) = T(15,3).

Also nonzero terms of A296508. - Omar E. Pol, Feb 11 2018

Conjecture 1: also widths of the successive terraces that we can find descending by the main diagonal of the pyramid described in A245092. Hence, bisection of A281012.

Conjecture 2: also number of central subparts in the symmetric representation of sigma of the numbers j that have the property that the number of parts in the symmetric representation of sigma(j) is odd.

Conjecture 3: Partial sums give A282131.