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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

Number of edges in the join of two complete graphs, each of order n, K_n * K_n. - Roberto E. Martinez II, Jan 07 2002

The power series expansion of the entropy function H(x) = (1+x)log(1+x) + (1-x)log(1-x) has 1/a_i as the coefficient of x^(2i) (the odd terms being zero). - Tommaso Toffoli (tt(AT)bu.edu), May 06 2002

Partial sums of A016813 (4n+1). Also with offset = 0, a(n) = (2n+1)(n+1) = A005408 * A000027 = 2n^2 + 3n + 1, i.e., a(0) = 1. - Jeremy Gardiner, Sep 29 2002

Sequence also gives the greatest semiperimeter of primitive Pythagorean triangles having inradius n-1. Such a triangle has consecutive longer sides, with short leg 2n-1, hypotenuse a(n) - (n-1) = A001844(n), and area (n-1)*a(n) = 6*A000330(n-1). - Lekraj Beedassy, Apr 23 2003

Number of divisors of 12^(n-1), i.e., A000005(A001021(n-1)). - Henry Bottomley, Oct 22 2001

More generally, if p1 and p2 are two arbitrarily chosen distinct primes then a(n) is the number of divisors of (p1^2*p2)^(n-1) or equivalently of any member of A054753^(n-1). - Ant King, Aug 29 2011

Number of standard tableaux of shape (2n-1,1,1) (n>=1). - Emeric Deutsch, May 30 2004

It is well known that for n>0, A014105(n) [0,3,10,21,...] is the first of 2n+1 consecutive integers such that the sum of the squares of the first n+1 such integers is equal to the sum of the squares of the last n; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2.

Less well known is that for n>1, a(n) [0,1,6,15,28,...] is the first of 2n consecutive integers such that sum of the squares of the first n such integers is equal to the sum of the squares of the last n-1 plus n^2; e.g., 15^2 + 16^2 + 17^2 = 19^2 + 20^2 + 3^2. - Charlie Marion, Dec 16 2006

a(n) is also a perfect number A000396 when n is an even superperfect number A061652. - Omar E. Pol, Sep 05 2008

Sequence found by reading the line from 0, in the direction 0, 6, ... and the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the generalized hexagonal numbers A000217. - Omar E. Pol, Jan 09 2009

For n>=1, 1/a(n) = Sum_{k=0..2*n-1} ((-1)^(k+1)*binomial(2*n-1,k)*binomial(2*n-1+k,k)*H(k)/(k+1)) with H(k) harmonic number of order k.

The number of possible distinct colorings of any 2 colors chosen from n colors of a square divided into quadrants. - Paul Cleary, Dec 21 2010

Central terms of the triangle in A051173. - Reinhard Zumkeller, Apr 23 2011

For n>0, a(n-1) is the number of triples (w,x,y) with all terms in {0,...,n} and max(|w-x|,|x-y|) = |w-y|. - Clark Kimberling, Jun 12 2012

a(n) is the number of positions of one domino in an even pyramidal board with base 2n. - César Eliud Lozada, Sep 26 2012

Partial sums give A002412. - Omar E. Pol, Jan 12 2013

Let a triangle have T(0,0) = 0 and T(r,c) = |r^2 - c^2|. The sum of the differences of the terms in row(n) and row(n-1) is a(n). - J. M. Bergot, Jun 17 2013

With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for A176230, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 11 2013

a(n) is the number of length 2n binary sequences that have exactly two 1's. a(2) = 6 because we have: {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}. The ordinary generating function with interpolated zeros is: (x^2 + 3*x^4)/(1-x^2)^3. - Geoffrey Critzer, Jan 02 2014

For n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n^(2*m) is a multiple of k + n is given by k = 2*n^(2*m) - n. - Derek Orr, Sep 04 2014

Binomial transform of (0, 1, 4, 0, 0, 0, ...) and second partial sum of (0, 1, 4, 4, 4, ...). - Gary W. Adamson, Oct 05 2015

a(n) also gives the dimension of the simple Lie algebras D_n, for n >= 4. - Wolfdieter Lang, Oct 21 2015

For n > 0, a(n) equals the number of compositions of n+11 into n parts avoiding parts 2, 3, 4. - Milan Janjic, Jan 07 2016

Also the number of minimum dominating sets and maximal irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Jun 29 and Aug 17 2017

As Beedassy's formula shows, this Hexagonal number sequence is the odd bisection of the Triangle number sequence. Both of these sequences are figurative number sequences. For A000384, a(n) can be found by multiplying its triangle number by its hexagonal number. For example let's use the number 153. 153 is said to be the 17th triangle number but is also said to be the 9th hexagonal number. Triangle(17) Hexagonal(9). 17*9=153. Because the Hexagonal number sequence is a subset of the Triangle number sequence, the Hexagonal number sequence will always have both a triangle number and a hexagonal number. n* (2*n-1) because (2*n-1) renders the triangle number. - Bruce J. Nicholson, Nov 05 2017

Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central valley and the largest Dyck path has a central peak, n >= 1. Thus all hexagonal numbers > 0 have middle divisors. (Cf. A237593.) - Omar E. Pol, Aug 28 2018

k^a(n-1) mod n = 1 for prime n and k=2..n-1. - Joseph M. Shunia, Feb 10 2019

Consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z: a(n+1) gives the semiperimeter of related triangles; A005408, A046092 and A001844 give the X, Y and Z values. - Ralf Steiner, Feb 25 2020

See A002939(n) = 2*a(n) for the corresponding perimeters. - M. F. Hasler, Mar 09 2020

It appears that these are the numbers k with the property that the smallest subpart in the symmetric representation of sigma(k) is 1. - Omar E. Pol, Aug 28 2021

The above conjecture is true. See A280851 for a proof. - Hartmut F. W. Hoft, Feb 02 2022

The n-th hexagonal number equals the sum of the n consecutive integers with the same parity starting at n; for example, 1, 2+4, 3+5+7, 4+6+8+10, etc. In general, the n-th 2k-gonal number is the sum of the n consecutive integers with the same parity starting at (k-2)*n - (k-3). When k = 1 and 2, this result generates the positive integers, A000027, and the squares, A000290, respectively. - Charlie Marion, Mar 02 2022

Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)} such that |A|=|B|=k and A+B={0,1,2,...,2*a(n)}} = 2*n. - Michael Chu, Mar 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 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

Gives an identity for sigma(n): alternating sum of row n equals the sum of divisors of n. For proof see Max Alekseyev link.

Row n has length A003056(n) hence column k starts in row A000217(k).

The number of positive terms in row n is A001227(n), the number of odd divisors of n.

If n = 2^j then the only positive integer in row n is T(n,1) = 2^(j+1) - 1.

If n is an odd prime then the only two positive integers in row n are T(n,1) = 2n - 1 and T(n,2) = n - 2.

If T(n,k) = 3 then T(n+1,k+1) = 1, the first element of the column k+1.

The partial sums of column k give the column k of A236104.

The connection with the symmetric representation of sigma is as follows: A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> A237270.

Alternating sum of row n equals the number of units cubes that protrude from the n-th level of the stepped pyramid described in A245092. - Omar E. Pol, Oct 28 2015

Conjecture: T(n,k) is the difference between the square of the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the square of the total number of partitions of all positive integers < n into exactly k consecutive parts. - Omar E. Pol, Feb 14 2018

From Omar E. Pol, Nov 24 2020: (Start)

T(n,k) is also the number of steps in the first n levels of the k-th double-staircase that has at least one step in the n-th level of the "Double- staircases" diagram, otherwise T(n,k) = 0, (see the Example section).

For the connection with A280851 see also the algorithm of A280850 and the conjecture of A296508. (End)

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

Apart from the alternating row sums and the sum of divisors function A000203 another connection with Euler's pentagonal theorem is that in the irregular triangle of A238442 the k-th column starts in the row that is the k-th generalized pentagonal number A001318(k) while here the k-th column starts in the row that is the k-th generalized hexagonal number A000217(k). Both A001318 and A000217 are successive members of the same family: the generalized polygonal numbers. - Omar E. Pol, Sep 23 2021

Other triangle with the same row lengths and alternating row sums equals sigma(n) is A252117. - Omar E. Pol, May 03 2022

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

This is a triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists successive blocks of k consecutive terms, where the m-th block starts with m, m>=1, and the first element of column k is in row k*(k+1)/2.

The partitions of n into consecutive parts are represented from the row n up to row A288529(n) as maximum, but in increasing order, exclusively in the columns where the blocks begin.

More precisely, the partition of n into exactly k consecutive parts (if such partition exists) is represented in the column k from the row n up to row n + k - 1 (see examples).

A288772(n) is the minimum number of rows that are required to represent in this table the partitions of all positive integers <= n into consecutive parts.

A288773(n) is the largest of all positive integers whose partitions into consecutive parts can be totally represented in the first n rows of this table.

A288774(n) is the largest positive integers whose partitions into consecutive parts can be totally represented in the first n rows of this table.

For a theorem related to this table see A286000.

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.