cp's OEIS Frontend

n-th row has length A001222(n) (n>1).

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

This is the original sequence of a large number of sequences connected with the section model of partitions.

Here "the n-th section of the set of partitions of any integer greater than or equal to n" (hence "the last section of the set of partitions of n") is defined to be the set formed by all parts that occur as a result of taking all partitions of n and then removing all parts of the partitions of n-1. For integers greater than 1 the structure of a section has two main areas: the head and tail. The head is formed by the partitions of n that do not contain 1 as a part. The tail is formed by A000041(n-1) partitions of 1. The set of partitions of n contains the sets of partitions of the previous numbers. The section model of partitions has several versions according with the ordering of the partitions or with the representation of the sections. In this sequence we use the ordering of A026791.

The section model of partitions can be interpreted as a table of partitions. See also A138121. - Omar E. Pol, Nov 18 2009

It appears that the versions of the model show an overlapping of sections and subsections of the numbers congruent to k mod m into parts >= m. For example:

First generation (the main table):

Table 1.0: Partitions of integers congruent to 0 mod 1 into parts >= 1.

Second generation:

Table 2.0: Partitions of integers congruent to 0 mod 2 into parts >= 2.

Table 2.1: Partitions of integers congruent to 1 mod 2 into parts >= 2.

Third generation:

Table 3.0: Partitions of integers congruent to 0 mod 3 into parts >= 3.

Table 3.1: Partitions of integers congruent to 1 mod 3 into parts >= 3.

Table 3.2: Partitions of integers congruent to 2 mod 3 into parts >= 3.

And so on.

Conjecture:

Let j and n be integers congruent to k mod m such that 0 <= k < m <= j < n. Let h=(n-j)/m. Consider only all partitions of n into parts >= m. Then remove every partition in which the parts of size m appears a number of times < h. Then remove h parts of size m in every partition. The rest are the partitions of j into parts >= m. (Note that in the section model, h is the number of sections or subsections removed), (Omar E. Pol, Dec 05 2010, Dec 06 2010).

Starting from the first row of triangle, it appears that the total numbers of parts of size k in k successive rows give the sequence A000041 (see A182703). - Omar E. Pol, Feb 22 2012

The last section of n contains A187219(n) regions (see A206437). - Omar E. Pol, Nov 04 2012

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

These are the squares of the entries of the triangle in A235791: T(n,k) = (A235791(n,k))^2.

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

Columns 1-3 (including the initial zeros) are A000290, A008794, A211547.

Also column k lists the partial sums of the k-th column of triangle A196020 which gives an identity for sigma.

Since all the elements of this sequence are squares, we can draw an illustration of the alternating sum of row n step by step, and a symmetric diagram for A000203, A024916, A004125; see example.

For more information about the diagram see A237593.

Since the prime factorization of 1 is the empty product (i.e., the multiplicative identity, 1), it follows that the prime signature of 1 is the empty multiset { }. (Cf. http://oeis.org/wiki/Prime_signature)

MathWorld wrongly defines the prime signature of 1 as {1}, which is actually the prime signature of primes.

The sequences A025487, A036035, A046523 consider the prime signatures of 1 and 2 to be distinct, implying { } for 1 and {1} for 2.

Since the prime signature of n is a partition of Omega(n), also true for Omega(1) = 0, the order of exponents is only a matter of convention (using reverse sorted lists of exponents would create a different sequence).

Here the multisets of nonzero exponents are sorted in increasing order; it is slightly more common to order them, as the parts of partitions, in decreasing order. This yields A212171. - M. F. Hasler, Oct 12 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

Same as A238130, with zeros omitted.

Last elements in rows are 1, 1, 2, 2, 1, 4, 2, 1, 6, 2, 1, 8, ... with g.f. -(x^6+x^4-2*x^2-x-1)/(x^6-2*x^3+1).

For n > 0, also the number of compositions of n with k + 1 runs. - Gus Wiseman, Apr 10 2020

Number of terms in n-th row is A001221(n) for n > 1.

From Reinhard Zumkeller, Aug 27 2011: (Start)

A008472(n) = Sum_{k=1..A001221(n)} T(n,k), n>1;

A007947(n) = Product_{k=1..A001221(n)} T(n,k);

A006530(n) = Max_{k=1..A001221(n)} T(n,k).

A020639(n) = Min_{k=1..A001221(n)} T(n,k).

(End)

Subsequence of A027750 that lists the divisors of n. - Michel Marcus, Oct 17 2015