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The "subparts" of the symmetric representation of sigma(n) are defined to be the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.

The number of layers of width 1 in the symmetric representation of sigma(n) is given in A250068.

The number of subparts in the first layer of the symmetric representation of sigma(n) is equal to A237271(n).

We can find the symmetric representation of sigma(n) as the terraces at the n-th level (starting from the top) of the stepped pyramid described in A245092.

(All above comments are essentially the same as the comments dated Nov 05 2016 at the old version of A275601, which was the same as A001227).

The sum of row n equals the number of subparts in the symmetric representation of sigma(n).

Conjecture:

The number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n.

From Hartmut F. W. Hoft, Dec 16 2016: (Start)

Proof:

Each row of the irregular triangle of A262045 can be interpreted as a step function of step sizes 1, 0, and -1. The numbers in row n are the widths of the segments in the parts of the symmetric representation of sigma(n). Each new subpart in a segment (in the left half) of row n starts at the same odd index that represents an odd divisor d of n in the irregular triangle of A237048. Either a subpart ends at an even index e, representing a second odd divisor, which satisfies d * e = oddpart(n), and thus the entire subpart is duplicated in the symmetric portion of the representation, or a subpart runs through the center and continues contiguously into the right half of the symmetric portion of the representation. In other words, the number of subparts in row n equals the number of odd divisors of n, i.e., the conjecture is true. (End)

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

This sequence is a companion to A279387 in which each contiguous sequence of identical widths w in A249223 are replaced by a single entry of w. Using the resulting distribution pattern of widths across all parts of the symmetric representation of sigma(n) the subparts at each level are counted in A279387.

The sequence of widths are computed first to the diagonal of the symmetric representation of sigma only for those numbers in set F defined in A341971. Then the reversed list less its first number is appended so that the width at the diagonal is not listed twice. Thus every row contains an odd number of entries and is symmetric about its center entry.

Let 1 <= n, 1 <= d <= s = A001227(n) and 1 <= k <= r = floor((sqrt(8*n + 1) - 1)/2). Let Q(n,d) be row n in the triangle of A341970, R(n,d) be row n in the triangle of A341970 and S(n,d) = R(n,Q(n,d)), then T(n,e) = S(n,e) for 1 <= e <= s and T(n,e) = S(n,2*s - e) for s < e <= 2*s - 1 is row n for this sequence.

For the construction of the n-th row of this triangle start with a copy of the n-th row of the triangle A196020.

Then replace each element of the m-th pair of positive integers (x, y) with the value (x - y)/2, where "y" is the m-th even-indexed term of the row, and "x" is its previous nearest odd-indexed term not used in another pair in the same row, if such a pair exist. Otherwise T(n,k) = A196020(n,k). (See example).

Observation 1: at least for the first 28 rows of the triangle the nonzero terms in the n-th row are also the subparts of the symmetric representation of sigma(n), assuming the ordering of the subparts in the same row does not matter.

Question 1: are always the nonzero terms of the n-th row the same as all the subparts of the symmetric representation of sigma(n)? If not, what is the index of the row in which appears the first counterexample?

Note that the "subparts" are the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.

For more information about "subparts" see A279387 and A237593.

About the question 1, it appears that the n-th row of the triangle A280851 and the n-th row of this triangle contain the same nonzero numbers, though in different order; checked through n = 250000. - Hartmut F. W. Hoft, Jan 31 2018

From Omar E. Pol, Feb 02 2018: (Start)

Observation 2: at least for the first 28 rows of the triangle we have that in the n-th row the odd-indexed terms, from left to right, together with the even-indexed terms, from right to left, form a finite sequence in which the nonzero terms are the same as the n-th row of triangle A280851, which lists the subparts of the symmetric representation of sigma(n).

Question 2: Are always the same for all rows? If not, what is the index of the row in which appears the first counterexample? (End)

Conjecture: the odd-indexed terms of the n-th row together with the even-indexed terms of the same row but listed in reverse order give the n-th row of triangle A296508 (this is the same conjecture from A296508). - Omar E. Pol, Apr 20 2018

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

Since the width of the single region of the symmetric representation of sigma( 2^ceiling((p-1)*(log_2 3) - 1) * 3^(p-1) ), for prime number p, at the diagonal equals p, this sequence contains an increasing subsequence (see A250071).

a(n) is also the number of layers of width 1 in the symmetric representation of sigma(n). For more information see A001227. - Omar E. Pol, Dec 13 2016

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

The sequence is the intersection of A239929 (sigma(j) has two parts) and of A241008 (sigma(j) has an even number of parts of width one).

The numbers in the sequence are precisely those defined by the formula for the triangle, see the link. The symmetric representation of sigma(j) has two parts, each part having width one, precisely when j = 2^(k - 1) * p where 2^k <= row(j) < p, p is prime and row(j) = floor((sqrt(8*j + 1) - 1)/2). Therefore, the sequence can be written naturally as a triangle as shown in the Example section.

The symmetric representation of sigma(j) = 2*j - 2 consists of two regions of width 1 that meet on the diagonal precisely when j = 2^(2^m - 1)*(2^(2^m) + 1) where 2^(2^m) + 1 is a Fermat prime (see A019434). This subsequence of numbers j is 3, 10, 136, 32896, 2147516416, ...[?]... (A191363).

The k-th column of the triangle starts in the row whose initial entry is the first prime larger than 2^(k+1) (that sequence of primes is A014210, except for 2).

Observation: at least the first 82 terms coincide with the numbers j with no middle divisors whose largest divisor <= sqrt(j) is a power of 2, or in other words, coincide with the intersection of A071561 and A365406. - Omar E. Pol, Oct 11 2023

The symmetric representation of sigma(m) has 3 regions of width 1 where the two extremal regions each have 2^k - 1 legs and the central region starts with the p-th leg of the associated Dyck path for sigma(m) precisely when m = 2^(k - 1) * p^2 where 2^k < p <= row(m), k >= 1, p >= 3 is prime and row(m) = floor((sqrt(8*m + 1) - 1)/2). Furthermore, the areas of the two outer regions are (2^k - 1)*(p^2 + 1)/2 each so that the area of the central region is (2^k - 1)*p; for a proof see the link.

Since the sequence is defined by a two-parameter expression it can be written naturally as a triangle as shown in the Example section.

A263951 is a subsequence of this sequence containing the squares of all those primes p for which the areas of the 3 regions in the symmetric representation of p^2 (p once and (p^2 + 1)/2 twice) are primes; i.e., p^2 and p^2 + 1 are semiprimes (see A070552). - Hartmut F. W. Hoft, Aug 06 2020