cp's OEIS Frontend

a(n) counts stacks of integer-length boards of total length n where no board overhangs the board underneath.

Number of graphical partitions on 2n nodes that contain a 1. E.g. a(3)=4 and so there are 4 graphical partitions of 6 that contain a 1, namely (111111), (21111), (2211) and (3111). Only (222) fails. - Jon Perry, Jul 25 2003

It would seem from Stanley that he regards a(0)=0 for this sequence and A001522. - Michael Somos, Feb 22 2015

In the article by Auluck is a typo in the formula (24), 2*Pi is missing in an exponent on the left side of the equation for Q(n). The correct term is exp(2*Pi*sqrt(n/3)), not just exp(sqrt(n/3)). - Vaclav Kotesovec, Jun 22 2015

A fountain is formed by starting with a row of coins, then stacking additional coins on top so that each new coin touches two in the previous row.

Also the number of Dyck paths for which the sum of the heights of the vertices that terminate an upstep (i.e., peaks and doublerises) is n. Example: a(4)=3 because we have UDUUDD, UUDDUD and UDUDUDUD. - Emeric Deutsch, Mar 22 2008

Also the number of ordered trees with path length n (follows from previous comment via a standard bijection). - Emeric Deutsch, Mar 22 2008

Probably first studied by Jim Propp (unpublished).

Number of compositions of n with c(1) = 1 and c(i+1) <= c(i) + 1. (Slide each row right 1/2 step relative to the row below, and count the columns.) - Franklin T. Adams-Watters, Nov 24 2009

With the additional requirement for weak unimodality one obtains A001524. - Joerg Arndt, Dec 09 2012

Also number of partitions of n with positive crank (n>=2), cf. A064391. - Vladeta Jovovic, Sep 30 2001

Number of smooth weakly unimodal compositions of n into positive parts such that the first and last part are 1 (smooth means that successive parts differ by at most one), see example. Dropping the requirement for unimodality gives A186085. - Joerg Arndt, Dec 09 2012

Number of weakly unimodal compositions of n where the maximal part m appears at least m times, see example. - Joerg Arndt, Jun 11 2013

Also weakly unimodal compositions of n with first part 1, maximal up-step 1, and no consecutive up-steps; see example. The smooth weakly unimodal compositions are recovered by shifting all rows above the bottom row to the left by one position with respect to the next lower row. - Joerg Arndt, Mar 30 2014

It would seem from Stanley that he regards a(0)=0 for this sequence and A001523. - Michael Somos, Feb 22 2015

From Gus Wiseman, Mar 30 2021: (Start)

Also the number of odd-length compositions of n with alternating parts strictly decreasing. These are finite odd-length sequences q of positive integers summing to n such that q(i) > q(i+2) for all possible i. The even-length version is A064428. For example, the a(1) = 1 through a(9) = 14 compositions are:

(1) (2) (3) (4) (5) (6) (7) (8) (9)

(211) (221) (231) (241) (251) (261)

(311) (312) (322) (332) (342)

(321) (331) (341) (351)

(411) (412) (413) (423)

(421) (422) (432)

(511) (431) (441)

(512) (513)

(521) (522)

(611) (531)

(612)

(621)

(711)

(32211)

(End)

In the Ferrers diagram of a partition x of n, count the dots in each diagonal parallel to the main diagonal (starting at the top-right, say). The result diag(x) is a smooth weakly unimodal composition of n into positive parts such that the first and last part are 1. For example, diag(5541) = 11233221. The function diag is many-to-one; the size of its codomain as a set is a(n). If diag(x) = diag(y), each hook of x can be slid by the same amount past the main diagonal to get y. For example, diag(5541) = diag(44331). - George Beck, Sep 26 2021

From Gus Wiseman, May 23 2022: (Start)

Conjecture: Also the number of integer partitions y of n with a fixed point y(i) = i. These partitions are ranked by A352827. The conjecture is stated at A238395, but Resta tells me he may not have had a proof. The a(1) = 1 through a(8) = 10 partitions are:

(1) (11) (111) (22) (32) (42) (52) (62)

(1111) (221) (222) (322) (422)

(11111) (321) (421) (521)

(2211) (2221) (2222)

(111111) (3211) (3221)

(22111) (4211)

(1111111) (22211)

(32111)

(221111)

(11111111)

Note that these are not the same partitions (compare A352827 to A352874), only the same count (apparently).

(End)

The above conjecture is true. See Section 4 of the Blecher-Knopfmacher paper in the Links section. - Jeremy Lovejoy, Sep 26 2022

A stack is formed by starting with a row of triangles pointing upwards, and then additional triangles pointing downwards can be inserted between adjacent triangles pointing upwards, and additional triangles pointing upwards can be put on top of triangles pointing downwards.

Number of compositions of n with a(1) = 1 and a(i+1) <= a(i) + 1 + mod(a(i),2).

From Peter Bala, Jul 12 2019: (Start)

These triangle stacks may be defined using Schröder paths. A Schröder path is a lattice path in the plane starting and ending on the x-axis, never going below the x-axis, using the steps (1,1) rise, (1,-1) fall or (2,0) flat. A small Schröder path is a Schröder path with no flat steps on the x-axis.

The area between a small Schröder path and the x-axis may be decomposed into a stack of unit area triangles of two types - up triangles with vertices at the lattice points (x, y), (x+1, y+1) and (x+2, y) and down triangles with vertices at the lattice points (x, y), (x-1, y+1) and (x+1, y+1). (End)

Number of partitions of n such that each part occurs at most twice, the largest part is unique and if k occurs as a part then all smaller positive integers occur.

Strongly unimodal compositions with first part 1 and each up-step is by at most 1 (left-smoothness); with this interpretation one should set a(0)=1; see example. Replacing "strongly" by "weakly" in the condition gives A001524. Dropping the requirement of unimodality gives A005169. [Joerg Arndt, Dec 09 2012]

Number of compositions of n where the first and the last parts are 1 and the absolute difference between consecutive parts is <=1 (smooth compositions).

Such a composition [c1,c2,c3,...] corresponds to a sandpile with c1(=1) grains in the first positions, c2 in the second, and so on. Assuming the critical slope is 1 (for the pile to be stable) we obtain the conditions on the compositions.

With the additional requirement of unimodality one gets A001522. [Joerg Arndt, Dec 09 2012]

Dropping the requirement that the first and last parts are 1 gives A034297. Restriction to weakly increasing (or decreasing) sums gives A034296. [Joerg Arndt, Jun 02 2013]

Also the number of compositions of n with first part 1, up-steps of at most 1, and no two consecutive up-steps. The sandpiles are recovered by shifting the rows above the bottom row to the left by one position relative to the next lower row. [Joerg Arndt, Mar 30 2014]

Also fountains of coins (cf. A005169) with no consecutive up-steps. Shift the top rows in the previous comment by half a position. [Joerg Arndt, Mar 30 2014]

Computed by R. K. Guy (see link).

This sequence is similar to various sequences with rows of coins (often pennies). Each brick must be offset by 1/2 brick from any bricks under it. However, a brick might only have a brick under one half, provided the pile is stable. A definition of stability is provided in the Paterson paper.

Since the main goal of the Paterson paper is the search for stable piles with maximum overhang (and not to find the number of stable piles), it is likely that some stable piles configurations not leading to significant overhangs have been passed over. So it is not easy to determine whether or not all the stable piles configurations have been accounted for! So even though I added graphic depictions for n=8 and n=9 (cf. A168368a), I will abstain for now from appending a(8) and a(9) to this sequence until it has been ascertained that none are missing (please verify A168368a).

Connected piles only (allowing piles with disconnected subpiles would produce a different sequence).

For this sequence I assumed that the bricks on any given row need not be contiguous. A different sequence would be produced if we required the bricks on any given row to be contiguous (fewer piles of bricks since the piles would constitute a subset of the ones obtained from the current piles).

a(0) is set to 0 (no pile) instead of 1 (empty pile) since the number a(n) of stable connected piles of n bricks with height <= 2 is F_n (n-th Fibonacci number) and at least 4 bricks are needed to make a stable connected pile of height greater than 2 and outgrow the Fibonacci sequence {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...}.

The number of stable connected piles of n bricks is T_n (n-th tribonacci numbers, with a(0)=0, a(1)=a(2)=1, T_n = {0, 1, 1, 2, 4, 7, 13, 24, 44, 81, ...}) up to n=5, after which it falls behind (for n >= 6).

The following few comments describe only one of many kinds of stacks known to be self stable or stabilizable via a balancing set on top, e.g. the n-diamonds stacks.

The n-diamonds (of width n, with n^2 bricks) are self stable (no balancing set needed) for n = 1 to 4 only (<1>, <2>, <3>, <4> only):

... & ..... & ....... & ...|=|...

... & ..... & ....... & ..|=|=|..

... & ..... & ..|=|.. & .|=|=|=|.

... & ..... & .|=|=|. & |=|=|=|=|

... & .|=|. & |=|=|=| & .|=|=|=|.

... & |=|=| & .|=|=|. & ..|=|=|..

|=| & .|=|. & ..|=|.. & ...|=|...

For n >= 5, the n-diamonds need a balancing set with at least 2^n-n^2-1 bricks on top of the n^2 bricks of the n-diamond for a combined total of 2^n-1 bricks.

For n=5 we need a balancing set of at least 2^5-5^2-1 = 6 bricks |B|, e.g., a 2-diamond <2> on top of |B|B| to stabilize <5>:

.<2>.

|B|B|

.<5>.

-----

= 6 = balancing bricks.

For n=6 we need a balancing set of at least 2^6-6^2-1 = 27 bricks |B| to stabilize <6>, e.g.:

.|B|.

|B|B|

.<2>. or .<2>.

|B|B| or |B|B|

.<2>. or .<2>. or .<2>. or .|B|.

|B|B| or |B|B| or |B|B| or |B|B|

.<2>. or .<2>. or .<3>. or .<2>. or .<3>.

|B|B| or |B|B| or |B|B| or |B|B| or |B|B|

.<2>. or .<3>. or .<3>. or .<4>. or .<4>.

|B|B| or |B|B| or |B|B| or |B|B| or |B|B|

.<6>. or .<6>. or .<6>. or .<6>. or .<6>.

----- .. ----- .. ----- .. ----- .. -----

= 27= .. = 29= .. = 28= .. = 27= .. = 29= balancing bricks.

For n=7 we need a balancing set of at least 2^7-7^2-1 = 78 bricks |B| to stabilize <7>, where the balancing set itself might need its own balancing subset as in the following example:

.|B|.

|B|B|

.<3>.

|B|B|

.<5>.

|B|B|

.<6>.

|B|B|

.<7>.

-----

= 79 = balancing bricks (where <5> itself needs 6 balancing bricks).