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From Wolfdieter Lang, Jun 26 2011: (Start)

This triangle S(n,m) appears as U_m(n) in the Knuth reference on p. 285. It is related to the Riordan triangle T_m(n) = A111125(n,m) by S(n,m) = A111125(n,m) - A111125(n-1,m), n >= m >= 1 (identity on p. 286).

Also, S(n,m)-S(n-1,m) = A111125(n-1,m-1), n >= 2, m >= 1 (identity on p. 286). (End)

These polynomials may be expressed in terms of the Faber polynomials of A263916 and are embedded in A127677 and A208513. - Tom Copeland, Nov 06 2015

Kekulé numbers for certain benzenoids.

T(n,k) is the number of Dyck (n+3)-paths with 3 peaks (UDs) and last descent of length k+1. For example, T(1,1)=3 counts UUDUDUDD, UDUUDUDD, UDUDUUDD. The number of Dyck n-paths containing k peaks and with last descent of length j is (j/n)*binomial(n,k-1)*binomial(n-j-1,k-2) (where as usual binomial(a,b)=0 for b < 0 except that binomial(-1,-1):=1). - David Callan, Jun 26 2006

As a rectangular array, this is the accumulation array (cf. A144112) of the rectangular array W given by w(i,j)=i+j-1; i.e., W=A002024 as a rectangular array. - Clark Kimberling, Sep 16 2008

T(n,k) gives the dimension of an irreducible representation of SU(3) whose Young diagram (n,k) has two rows of length n and k, respectively. - Dimitris Cardaris, May 10 2025

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

Second derivative of the n-th Chebyshev polynomial (of the first kind) evaluated at x=1.

The second derivative at x=-1 is just (-1)^n * a(n).

The difference between two consecutive terms generates the sequence a(n+1) - a(n) = A002492(n).

Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, |q-p| and |q-p|. - Wesley Ivan Hurt, Apr 15 2018

Number of different sizes occurring among the A002415(n) = n^2*(n^2-1)/12 squares that can be drawn using points of an n X n square array as corners.

a(n) is also the number of rectangular isosceles triangles, distinct up to congruence, on an n X n grid (or geoboard). - Martin Renner, May 03 2011

Mirror image of triangle A090181, another version of triangle of Narayana (A001263).

Equals A133336*A130595 as infinite lower triangular matrices. - Philippe Deléham, Oct 23 2007

A triangulation is simple if it contains no separating 3-cycle. There are n interior nodes and m+3 nodes on the boundary. - Andrew Howroyd, Feb 24 2021

Sascha Kurz has proved that one can assume that the points belong to the square grid Z X Z.

So we obtain the same values if we replace the definition by: a(n) is the maximal number of squares that can be formed from n points chosen from the infinite square grid.

In other words, take the infinite square grid. Pick a set S of n grid points, and let c(S) be the number of subsets of four points of S that form a square of any (nonzero) size. Then a(n) = maximum of c(S) over all choices of S.

The general problem of estimating the maximal number of similar figures within point-configurations was studied in [Elekes--Erdos]. References can be found in [Matousek, p. 47] and [AFKK]. Note that the present problem concerning squares shares several properties with the case of right isosceles triangles treated in [AFR].

The following remarks are out of date, and will be revised soon to reflect progress made in October 2021 by a number of members of the Sequence Fans Mailing List.

A more detailed account will be found in the report by Sascha Kurz et al. which is nearing completion.

[Comments revised to this point by N. J. A. Sloane, Nov 02 2021]

Values for n <= 9 [now 16] are exact and are the same for a and b (see proofs below by Hugo van der Sanden and Benoît Jubin for n <= 7 and Sascha Kurz for n <= 9, which furthermore classify all optimal configurations for these values). Values for n > 9 are conjectural since they were obtained by exhaustive search for grid points within a square of side ceil(sqrt(n)), which is a reasonable assumption. A proof that optimal configurations (with gcd of side lengths equal to 1) have a diameter at most f(n) with f of moderate growth would permit exact computation of values from exhaustive searches.

Asymptotic behavior:

One has a(n), b(n) = Theta(n^2):

Upper bound: Since two vertices determine squares, one has b(n) = O(n^2). More explicitly: a pair of points uniquely determines the square that has it as diagonal, and a square has two diagonals, so b(n) <= n(n-1)/4 ~ n^2/4.

Lower bound: When n = m^2, the set of all grid points in [0, m-1]^2 yields S = n(n-1)/12 squares. Indeed, for a in [0, m-1] and b in [1, m], the square formed on (a,0) and (0,b) (as its "lower-left side") has other vertices (a+b,b) and (a,a+b), so there are (m-(a+b))^2 translates of that square. Therefore, S = Sum_{a=0..m-1} Sum_{b=1..m} (m - (a + b))^2. By change of summation indices ((a,c) := (a,a+b)), expanding, using the sum of the first m integers, squares, cubes, and refactoring, one obtains S = n(n-1)/12. Since a is nondecreasing, one has a(n) >= (n-1)(n-2)/12 ~ n^2/12.

We actually have better lower and upper bounds:

0.09... = (1-2/Pi)/4 <= liminf a(n)/n^2 <= limsup b(n)/n^2 <= 1/6 = 0.16...

The upper bound is given in [AFR] (which counts isosceles right triangles, so their value has to be divided by four, the number of isosceles right triangles formed on a square). This gives b(n) <= (2/3*(n - 1)^2 - 5/3)/4.

Lower bound (due to Peter Munn, see SeqFan post in the links): For r >= 0, denote by D(r) the disc centered at the origin with radius r. If A is a point on the boundary of D(r), then the set of points B such that the square with diagonal AB is included in D(r) is a lens-shaped region of area (Pi-2)r^2 (as a proportion of the disc area, this is twice A258146). Therefore, the number S of grid-squares included in D(R) can be estimated as follows: since the set of squares with at least two vertices equidistant from the origin is negligible, we can assume that every square has a unique vertex furthest from the origin, say at distance r, which corresponds to the A above. Then the opposite vertex B is in the region computed above, and the L^1 (aka rectilinear, or Manhattan) distance between A and B is even (being opposite vertices, they are two sides apart), so we have to divide that area by 2. There are approximately 2*Pi*r grid points at a distance approximately r from the origin, so at first order, S ~= Integral_{r=0..R} Pi*r(Pi - 2)r^2 dr = Pi(Pi-2)/4 R^4. Since the disc D(R) contains approximately Pi*R^2 points, one obtains S ~= (1-2/Pi)/4 n^2.

Conjecture: the asymptotic density of the numbers n such that there is no maximal arrangement formed by all the grid points within a suitably chosen circle, is 0. - Peter Munn, Sep 30 2021

For the known values of a(n) (n <= 17), there is a maximal arrangement formed using circles as specified by the conjecture above. For n <= 100 no arrangement has yet been found to contain more squares than the best attained using a circle as specified by A348469. - Peter Munn, Nov 12 2021

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.

The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).