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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.

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 in decreasing order, where the m-th block starts with k + m - 1, m>=1, and the first element of column k is in the 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, 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.

Theorem: the smallest part of the partition of n into exactly k consecutive parts (if such partition exists) equals the number of positive integers <= n having a partition into exactly k consecutive parts.

In the triangle the first partition with m parts appears as the last partition in row A000217(m), m >= 1. - Omar E. Pol, Mar 23 2022

For m >= 0, row 2^m consists of just one element (2^m). - Paolo Xausa, May 24 2025

Conjecture 1: T(n,k) is the number of parts in the partition of n into k consecutive parts, if T(n,k) > 0.

Conjecture 2: row sums give A204217, which should be also the total number of parts in all partitions of n into consecutive parts.

(The conjectures are true. See Joerg Arndt's proof in the Links section.) - Omar E. Pol, Jun 14 2017

From Omar E. Pol, May 05 2020: (Start)

Theorem: Let T(n,k) be an irregular triangle read by rows in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in the row that is the k-th (m+2)-gonal number, with n >= 1, k >= 1, m >= 0. T(n,k) is also the number of parts in the partition of n into k consecutive parts that differ by m, including n as a valid partition. Hence the sum of row n gives the total number of parts in all partitions of n into consecutive parts that differ by m.

About the above theorem, this is the case for m = 1. For m = 0 see the triangle A127093, in which row sums give A000203. For m = 2 see the triangle A330466, in which row sums give A066839 (conjectured). For m = 3 see the triangle A330888, in which row sums give A330889.

Note that there are infinitely many triangles of this kind, with m >= 0. Also, every triangle can be represented with a diagram of overlapping curves, in which every column of triangle is represented by a periodic curve. (End)

a(n) is twice the number of partitions of n into consecutive parts, minus the number of partitions of n into distinct parts such that the greatest part equals the number of all parts.

For a visualization of the sequence, consider a diagram formed for infinitely many double staircases as shown in the Example section.

a(n) is the number of horizontal line segments (or steps) that are only in the n-th level of the structure, starting from the top.

The total length of all vertical line segments that are adjacent and below the steps of the n-th level of the structure equals twice the total number of parts in all partitions of n into consecutive parts.

Note that in the n-th double staircase the top step is located in the level A000217(n), n >= 1, every horizontal line segment has length 1, and every vertical line segment has length n.

a(n) is also the number of horizontal line segments of length 1 or 2 in the n-th level of the similar diagram used to represent the sequence A237593 and other isosceles triangles related to A237593.

a(n) is odd if and only if n is a nonzero triangular number (A000217).

Double-staircases theorem of the sum of divisors: the total number of steps from level n to the top of all the odd-indexed double staircases that have at least one step in the level n, minus the total number of steps from the level n to the top of all the even-indexed double staircases that have at least one step in the level n equals sigma(n) = A000203(n).

The above theorem shows a symmetry of sigma in accordance with the symmetric Dyck Paths described in A237593 and with the pyramid described in A245092.

For the connection with the partitions into consecutive integers see also A196020, since we can see here that A196020(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 diagram, otherwise A196020(n,k) = 0. Also, it is the width of the mentioned staircase in n-th level of the diagram.

It appears that odd primes (A065091) are also the levels where there are steps in the staircases 1 and 2, but no step from other staircases.

It appears that powers of 2 (A000079) are also the levels where there are only one or two steps in total.

This sequence could be related to several other sequences (see the Crossrefs section of A262626).

Conjecture 1: T(n,k) is the largest part of the partition of n into k consecutive parts, if T(n,k) > 0.

Conjecture 2: row sums give A286015.

Trapezoidal interpretation from Peter Munn, Jun 18 2017: (Start)

There is one to one correspondence between nonzero T(n,k) and trapezoidal area patterns of n dots on a triangular grid, if we include the limiting cases of triangular patterns, straight lines (k=1) or a single dot (k=n=1). The corresponding pattern has T(n,k) dots in its longest side, k dots in the two adjacent sides and T(n,k)-k+1 dots in the fourth side (where a count of 1 dot may be understood as signifying that side's absence).

Reason: From the definition, for k >= 1, m >= 0, T(A000217(k)+km,k) = k+m, where A000217(k) = k(k+1)/2, the k-th triangular number. First element of column k is T(A000217(k),k) = k: this matches a triangular pattern of A000217(k) dots with 3 sides of k dots. Looking at this pattern as k rows of 1..k dots, extend each row by m dots to create a trapezoidal pattern of A000217(k)+km dots with a longest side of k+m dots and adjacent sides of k dots: this matches T(A000217(k)+km,k) = k+m. As nonzero elements in column k occur at intervals of k, every nonzero T(n,k) has a match. Every trapezoidal pattern can be produced by extending a triangular pattern as described, so they all have a match.

The truth of conjecture 1 follows, since each nonzero T(n,k) = k+m corresponds to a trapezoidal pattern of n dots having k rows with lengths (1+m)..(k+m).

The A270877 sieve is related to this sequence because it eliminates n if it is the sum of consecutive numbers whose largest term has survived the sifting (which may likewise be seen in terms of a trapezoidal dot pattern and its longest side). So the sieve eliminates n if any lesser numbers in A270877 are in row n of this sequence.

(End)

For m >= 0, row 2^m consists of just one element (2^m). - Paolo Xausa, May 24 2025

The one-part partition n = n is included in the count.

For the relation to pentagonal numbers see also A330888.

To partition n into k parts, we see if m exists such that m + (m + 1) + ... + (m + k - 1) = k*m + binomial(k, 2) = n exists. a(n) = 1 if and only if (n - binomial(k, 2)) / k is an integer and larger than 0. - David A. Corneth, Apr 28 2017

It appears that this a full version of the irregular triangle A237048. - Omar E. Pol, Apr 28 2017

The value of a(n) can never exceed 1, since that would imply the existence of distinct equal-length ranges of consecutive integers that add up to the same number, which is impossible. - Sidney Cadot, Jan 22 2023

The one-part partition n = n is included in the count.

For the relation to the octagonal numbers see also A334947.