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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

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)

Also the rows of both triangles A237270 and A237593 interleaved.

Also, irregular triangle read by rows in which T(n,k) is the area of the k-th region (from left to right in ascending diagonal) of the n-th symmetric set of regions (from the top to the bottom in descending diagonal) in the two-dimensional diagram of the perspective view of the infinite stepped pyramid described in A245092 (see the diagram in the Links section).

The diagram of the symmetric representation of sigma is also the top view of the pyramid, see Links section. For more information about the diagram see also A237593 and A237270.

The number of cubes at the n-th level is also A024916(n), the sum of all divisors of all positive integers <= n.

Note that this pyramid is also a quarter of the pyramid described in A244050. Both pyramids have infinitely many levels.

Odd-indexed rows are also the rows of the irregular triangle A237270.

Even-indexed rows are also the rows of the triangle A237593.

Lengths of the odd-indexed rows are in A237271.

Lengths of the even-indexed rows give 2*A003056.

Row sums of the odd-indexed rows gives A000203, the sum of divisors function.

Row sums of the even-indexed rows give the positive even numbers (see A005843).

Row sums give A245092.

From the front view of the stepped pyramid emerges a geometric pattern which is related to A001227, the number of odd divisors of the positive integers.

The connection with the odd divisors of the positive integers is as follows: A261697 --> A261699 --> A237048 --> A235791 --> A237591 --> A237593 --> A237270 --> this sequence.

Conjecture: the positive terms in row n are the odd divisors of n.

Note that the elements appear with an unusual ordering, for example; row 45 is 1, 45, 3, 0, 5, 15, 0, 0, 9.

The positive terms give A261697.

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 sum of row n is A000593(n).

The connection with the symmetric representation of sigma is as follows: A237048 --> A235791 --> A237591 --> A237593.

Proof of the conjecture: let n = 2^m*s*t with s and t odd. The property stated in A237048 verifies the conjecture with odd divisor k <= A003056(n) of n in position k and odd divisor t > A003056(n) in position 2^(m+1)*s. Therefore reading in row n the nonzero odd positions from left to right and then the nonzero even positions from right to left gives all odd divisors of n in increasing order. - Hartmut F. W. Hoft, Oct 25 2015

A237048 gives the signum function (A057427) of this sequence. - Omar E. Pol, Nov 14 2016

From Peter Munn, Jul 30 2017: (Start)

Each odd divisor d of n corresponds to n written as a sum of consecutive integers (n/d - (d-1)/2) .. (n/d + (d-1)/2). After canceling any corresponding negative and positive terms and deleting any zero term, the lower bound becomes abs(n/d - d/2) + 1/2, leaving k terms where k = n/d + d/2 - abs(n/d - d/2). It can be shown T(n,k) = d.

This sequence thereby defines a one to one relationship between odd divisors of n and partitions of n into k consecutive parts.

The relationship is expressed below using 4 sequences (with matching row lengths), starting with this one:

A261699(n,k) = d, the odd divisor.

A211343(n,k) = abs(n/d - d/2) + 1/2, smallest part.

A285914(n,k) = k, number of parts.

A286013(n,k) = n/d + (d-1)/2, largest part.

If no partition of n into k consecutive parts exists, the corresponding sequence terms are 0.

(End)

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

Row n is a permutation of the n-th row of A237591 for some n, hence the sequence is a permutation of A237591.

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

Row n is a permutation of the n-th row of A237591 for some n, hence the sequence is a permutation of A237591.

Row n gives the last A001227(n) terms of the n-th row of A379630 and of A379631.

For a correspondence between the row n and the partitions of n into consecutive parts see A379630.

n is an odd prime if and only if a(n) = 1 + a(n-1) and A237591(n,k) = A237591(n-1,k) for the values of k distinct of 2.

For k > 1 there are five numbers k in the sequence.

For more information see A237593.