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All entries in the triangle are nonnegative since the number of 1's in odd-numbered columns of A237048 prior to column j, 1 <= j <= row(n), is at least as large as the number of 1's in even-numbered columns through column j. As a consequence:

(a) The two adjacent symmetric Dyck paths whose legs are defined by adjacent rows of triangle A237593 never cross each other (see also A235791 and A237591) and the rows in this triangle describe the widths between the legs.

(b) Let legs(n) denote the n-th row of triangle A237591, widths(n) the n-th row of this triangle, and c(n) the rightmost entry in the n-th row of this triangle (center of the Dyck path). Then area(n) = 2 * legs(n) . width(n) - c(n), where "." is the inner product, is the area between the two adjacent symmetric Dyck paths.

(c) For certain sequences of integers, it is known that area(n) = sigma(n); see A238443, A245685, A246955, A246956 and A247687.

Right border gives A067742. - Omar E. Pol, Jan 21 2017

For a proof that T(n, k) = |{ d : d|n and k/2 < d <= k }|, for 1 <= k <= row(n), where row(n) is the length of row n, an identity suggested by Peter Munn, see the link. A corollary to it is that the number of divisors of n in the half-open interval (row(n)/2, row(n)] equals the width of the symmetric representation of n at the diagonal: T(n, row(n)) = | { d : d|n and row(n)/2 < d <= row(n) } |. See also the comments and conjectures of Michel Marcus in A067742 and A237593. - Hartmut F. W. Hoft, Jun 24 2024

From Omar E. Pol, Jul 24 2024: (Start)

Conjecture 1: Every column is a periodic sequence.

Conjecture 2: The periods of the columns 1..8 are respectively: 1, 2, 6, 12, 60, 60, 420, 840.

Question 1: Is the period of the column k equal to A003418(k)? (End).

From Omar E. Pol, Jul 26 2024: (Start)

Column 1 gives A000012.

Column 2 gives A000035.

Conjecture 3: Column 3 gives [2, 0] together with A115357, hence column 3 gives 2 together with A171182.

Question 2: Except the first nine terms of A337976, is the column 4 the same as A337976?

Question 3: Except the first 14 terms of A366981, is the column 5 the same as A366981? (End)

From Hartmut F. W. Hoft, Aug 01 2024: (Start)

Conjectures 1 and 2 are true and the answer to question 1 is affirmative.

By definition, each column k in triangle T237048(n, k) of sequence A237048 is a periodic sequence of period k. Since the k-th term in row n of the triangle T(n, k) = Sum_{i=1..k} (-1)^(i+1) * T237048(n, i), with 1 <= k <= A003056(n), each initial subsequence T(n, 1) .. T(n, k) of row n in this triangle is periodic of period lcm(1, .. , k) = A003418(k). This implies that each column k in this sequence has period A003418(k).

Conjecture 3 and Question 2 are true. Since T237048(n, 1) = 1, T237208(n, 2) = 1 if n odd and 0 if n even, T237048(n, 3) = 1 if 3|n and 0 otherwise, and T237048(n, 4) = 1 if 4|(n-2) and 0 otherwise, equations T249223(n, 3) = 1 - (n mod 2) + delta( n mod 3) and T249223(n, 4) = 1 - (n mod 2) + delta( n mod 3) - delta( (n-2) mod 4) hold where delta(k) = 1 if k = 0 and 0 otherwise. With the 3rd column starting at n = A000217(3) = 6, each period starting in a row that is a multiple of 6 is [ 2 0 1 1 1 0 ], and appropriate shifts yield A115357 and A171182. With the 4th column starting at n = A000217(4) = 10, each period starting in a row n with 12|(n+2) is [ 0 0 2 0 0 1 1 0 1 0 1 1 ], and with a shift of 9 yields the apparently periodic A337976(10), A337976(11), ... (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

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

The 26 entries starting with a(2) = 6 are products of powers of consecutive primes starting with 2, except for a(12) = 32760 and a(15) = 98280 (which are missing 11), and a(26) = 942480 (which is missing 13).

a(n) is the smallest number k such that the symmetric representation of sigma(k) has n layers. For more information see A279387. - Omar E. Pol, Dec 16 2016

Row 1 of A253258. - Omar E. Pol, Apr 15 2018

From Hartmut F. W. Hoft, Jun 10 2024: (Start)

All terms a(n) <= 1.75*10^7 have a symmetric representation of sigma that consists of a single part and they are abundant for n > 2. Numbers a(1) = 1, a(2) = 6, and a(4) = 120 are unimodal while numbers a(6) = 840, a(14) = 50400, a(18) = 138600, a(24) = 960960, a(26) = 942480, a(32) = 2827440, a(44) = 8648640 have a single extent of maximum width, but are not unimodal.

Conjecture: The symmetric representation of sigma for every term consists of a single part and it is unimodal only for a(1), a(2), and a(4).

As a consequence, this sequence would be a subsequence of A174973, and all a(n), n > 2, would be abundant. (End)

The first eight entries in A071561 but not in this sequence are 75, 78, 102, 105, 114, 138, 174 & 175.

The first eight entries in A239929 but not in this sequence are 21, 27, 33, 39, 51, 55, 57 & 65.

The union of this sequence and A241010 equals A174905 (see link in A174905 for a proof). Updated by Hartmut F. W. Hoft, Jul 02 2015

Let n = 2^m * Product_{i=1..k} p_i^e_i = 2^m * q with m >= 0, k >= 0, 2 < p_1 < ... < p_k primes and e_i >= 1, for all 1 <= i <= k. For each number n in this sequence k > 0, at least one e_i is odd, and for any two odd divisors f < g of n, 2^(m+1) * f < g. Let the odd divisors of n be 1 = d_1 < ... < d_2x = q where 2x = sigma_0(q). The z-th region of the symmetric spectrum of n has area a_z = 1/2 * (2^(m+1) - 1) *(d_z + d_(2x+1-z)), for 1 <= z <= 2x. Therefore, the sum of the area of the regions equals sigma(n). For a proof see Theorem 6 in the link of A071561. - Hartmut F. W. Hoft, Sep 09 2015, Sep 04 2018

First differs from A071561 at a(43). - Omar E. Pol, Oct 06 2018

Let k be a nonnegative integer such that F(k) = 2^(2^k) + 1 is prime (a Fermat prime A019434), then m = (F(k)-1)*F(k)/2 appears in the sequence.

Conjecture: a(1)=3 is the only odd term of the sequence.

Conjecture: All terms of the sequence are of the above form derived from Fermat primes.

The sequence has 5 (known) terms in common with sequences A055708 (k-1 | sigma(k)) and A056006 (k | sigma(k)+2) since {a(n)} is a subsequence of both.

The first five terms of the sequence are respectively congruent to 3, 4, 4, 4, 4 modulo 6.

After a(5) there are no further terms < 8*10^9.

Up to m = 1312*10^8 there are no further terms in the class congruent to 4 modulo 6.

a(6) > 10^12. - Donovan Johnson, Dec 08 2011

a(6) > 10^13. - Giovanni Resta, Mar 29 2013

a(6) > 10^18. - Hiroaki Yamanouchi, Aug 21 2018

See A125246 for numbers with deficiency 4, i.e., sigma(m) = 2*m - 4, and A141548 for numbers with deficiency 6. - M. F. Hasler, Jun 29 2016 and Jul 17 2016

A term m of this sequence multiplied by a prime p not dividing it is abundant if and only if p < m-1. For each of a(2..5) there is such a prime near this limit (here: 7, 127, 30197, 2147483647) such that a(k)*p is a primitive weird number, cf. A002975. - M. F. Hasler, Jul 19 2016

Any term m of this sequence can be combined with any term j of A088831 to satisfy the property (sigma(m) + sigma(j))/(m+j) = 2, which is a necessary (but not sufficient) condition for two numbers to be amicable. [Proof: If m = a(n) and j = A088831(k), then sigma(m) = 2m-2 and sigma(j) = 2j+2. Thus, sigma(m) + sigma(j) = (2m-2) + (2j+2) = 2m + 2j = 2(m+j), which implies that (sigma(m) + sigma(j))/(m+j) = 2(m+j)/(m+j) = 2.] - Timothy L. Tiffin, Sep 13 2016

At least the first five terms are a subsequence of A295296 and of A295298. - David A. Corneth, Antti Karttunen, Nov 26 2017

Conjectures: all terms are second hexagonal numbers (A014105). There are no terms with middle divisors. - Omar E. Pol, Oct 31 2018

The symmetric representation of sigma(m) of each of the 5 numbers in the sequence consists of 2 parts of width 1 that meet at the diagonal (subsequence of A246955). - Hartmut F. W. Hoft, Mar 04 2022

The first five terms coincide with the sum of two successive terms of A058891. The same is not true for a(6), if such exists. - Omar E. Pol, Mar 03 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

The symmetric representation of sigma(k) has nondecreasing width to the diagonal precisely when all odd divisors counted in the k-th row of A237048 occur at odd indices. If we write k = 2^m * q with m >= 0 and q odd, this property is equivalent to q < 2^(m+1).

The values for a(11), a(13), a(17) and a(19) were computed directly using the formula k = 2^m * 3^(p-1) where p is one of the four primes and m the smallest exponent so that 3^(p-1) < 2^(m+1). Each of these numbers has a symmetric representation of nondecreasing width ending in a prime number width, and they are the first such numbers since the number of divisors of an odd number is a prime precisely when the number is a power of a prime.

The other numbers listed whose symmetric representations of sigma(k) have nondecreasing width are smaller than 7500000. The only additional numbers k <= 100000000 are a(24) = 7096320, a(27) = 90316800 and a(32) = 85155840.

See A340506 for another way to look at this data. - N. J. A. Sloane, Jan 23 2021

The first eight entries in A071562 but not in this sequence are 6, 12, 15, 18, 20, 24, 28 & 30.

The first eight entries in A238443 but not in this sequence are 6, 12, 18, 20, 24, 28, 30 & 36.

The union of A241008 and of this sequence equals A174905 (for a proof see link in A174905).

Let n = 2^m * product(p_i^e_i, i=1,...,k) = 2^m * q with m >= 0, k >= 0, 2 < p_1, ...< p_k primes and e_i >= 1, for all 1 <= i <= k. For each number n in this sequence all e_i are even, and for any two odd divisors f < g of n, 2^(m+1) * f < g. The sum of the areas of the regions r(n, z) equals sigma(n). For a proof of the characterization and the formula see the theorem in the link below.

Numbers 3025 = 5^2 * 11^2 and 510050 = 2^1 * 5^2 * 101^2 are the smallest odd and even numbers, respectively, in the sequence with two distinct odd prime factors.

Among the 706 numbers in the sequence less than 1000000 (see link to the table) there are 143 that have two different odd prime factors, but none with three. All numbers with three different odd prime factors are larger than 500000000. Number 4450891225 = 5^2 * 11^2 * 1213^2 is in the sequence, but may not be the smallest one with three different odd prime factors. Note that 1213 is the first prime that extends the list of divisors of 3025 while preserving the property for numbers in this sequence.

The subsequence of numbers n = 2^(k-1) * p^2 satisfying the constraints above is A247687.

n = 3^(2*h) = 9^h = A001019(h), h>=0, is the smallest number for which the symmetric representation of sigma(n) has 2*h+1 regions of width one, for example for h = 1, 2, 3 and 5, but not for h = 4 in which case 3025 = 5^2 * 11^2 < 3^8 = 6561 is the smallest (see A318843). [Comment corrected by Hartmut F. W. Hoft, Sep 04 2018]

Computations using this characterization are more efficient than those of Dyck paths for the symmetric representations of sigma(n), e.g., the Mathematica code below.

What do the parts and subparts of the symmetric representation of sigma(n) represent? (cf. A237270, A280851). This sequence gives an answer.

To calculate a(n) we must follow a geometric algorithm composed of three stages as follows:

Stage 1 (Construction):

On the infinite square grid we draw the diagram called "double-staircases" with n levels described in A335616.

Then we label the double-staircases from left to right starting from k = 1 to the central column of the diagram.

Note that k is also the difference in height between two steps of the ladder it would have if we drew it with more than one step.

Stage 2 (Debugging):

We remove all double-staircases that do not have at least one step at the level 1 of the diagram starting from the base.

Now the number of steps in the k-th double-staircase is equal to A196020(n,k).

Stage 3 (Annihilation):

From left to right, we remove each even-indexed double-staircase along with the steps of the nearest odd-indexed double-staircase that are just above it.

As a result of this geometric algorithm a diagram is obtained which can have only double-staircases, only simple-staircases, or both at the same time.

We call double-staircases those that have a step in the central column of the diagram. The rest are simple-staircases forming one or more symmetrical pairs equidistant from the central column of the diagram.

We call "parts" of the diagram to the staircases (and the cells below them) that are separated from each other by columns of zero height.

We call "subparts" of the diagram to the polygons formed by the cells that are under the staircases.

a(n) is the total area (or the total number of cells) under all the staircases, with multiplicity.

The connection with the polygonal numbers is as follows:

The area under a double-staircase labeled with the number k is equal to the m-th (k+2)-gonal number plus the (m-1)-th (k+2)-gonal number, where m is the number of steps on one side of the ladder from the base to the top.

If k = 1 then the area under the double-staircase is also equal to n^2 = A000290(n).

The area under a simple-staircase labeled with the number k is equal to the m-th (k+2)-gonal number, where m is the number of steps.

If n is a power of 2, or if n is an odd prime number, or if n is an even perfect number, or if n is a member of A246955, then the calculation of a(n) is easy (see the Formula section).

The connection with the symmetric representation of sigma(n) or "SRS(n)" is as follows:

The total number of steps in the diagram is equal to A000203(n), equaling the total area (or the number of cells) in the SRS(n).

The number of parts in the diagram is equal to A237271(n) equaling the number of parts in the SRS(n).

The number of double-staircases (also the number of steps in the central column in the diagram) is equal to A067742(n), equaling the number of central subparts in the SRS(n).

The number of simple-staircases is equal to A281009(n), equaling the total number of equidistant subparts in the SRS(n).

The total number of staircases is equal to A001227(n), equaling the number of subparts in the SRS(n).

The number of columns in the diagram is equal to 2*n - 1, equaling the number of "widths" in the SRS(n) (cf. A249351).

The number of steps in the successive parts of the diagram gives the n-th row of triangle A237270, equaling the successive parts in the SRS(n).

The number of steps in the successive staircases from left to right gives the n-th row of triangle A280851, equaling the successive subparts from left to right in the SRS(n).

The diagram is essentially the front view of a three-dimensional structure whose base is the symmetric representation of sigma(n), so a(n) is also the total number of cubic cells (or cubes) in the structure.

The number of polycubes in the structure is equal to A237271(n), equaling the number of parts in the SRS(n).

For some values of n the three-dimensional structure resembles a "ziggurat" which is a type of massive structure built in ancient Mesopotamia.

It appears that the geometric algorithm described here is equivalent to the numerical algorithm conjectured in A280850 in the sense that both allow the value of the subparts of the SRS(n) to be computed.

Both algorithms would also be equivalent to the geometric transformation of the isosceles triangle described in A237593 which, when folded, transforms into the vertical faces of the structure of the stepped pyramid described in A245092.

Note that the stage 2 (Debugging) is in correspondence with the formula A196020(n,k) = A237048(n,k)*A338721(n,k) which transform the triangle A338721 into the triangle A196020. - Omar E. Pol, Jun 23 2022