OFFSET

1,3

COMMENTS

The outer diagonal is A059435.

The second outer diagonal is A090442.

The third outer diagonal is essentially 2*A068766.

The first column is A011782.

The second column is essentially A057711 (not considering its first two terms).

The second column is essentially A129952 (not considering its first two terms).

The second column is essentially 2*A001792.

The differences between the terms of the second column is essentially 2*A045623.

The third column is essentially 4*A084266.

The cumulative sums of the third column are essentially 4*A176027.

T(n,k) = 0 for n < k. If this overriding constraint is not applied, you get A059576. - Franklin T. Adams-Watters, Jul 24 2011

For n >= 2 and 1 <= k <= n, T(n,k) is the number of bimonotone subdivisions of a 2-row grid with n points on the first row and k points on the second row (with the lower left point of the grid being the origin). A bimonotone subdivision of a convex polygon (the convex hull of the grid) is one where the internal dividing lines have nonnegative (including infinite) slopes. See Robeva and Sun (2020). - Petros Hadjicostas and Michel Marcus, Jul 15 2020

LINKS

Andrea Raffetti, Rows n = 1..13 of triangle

Elina Robeva and Melinda Sun, Bimonotone Subdivisions of Point Configurations in the Plane, arXiv:2007.00877 [math.CO], 2020.

FORMULA

T(n,1) = 2^(n-2) for n >= 2.

T(n,2) = n*2^(n-2) for n >= 2.

T(n,3) = 2^(n-2)*((n-k+1)^2 + 7*(n-k+1) + 4)/2 = 2^(n-3)*(n^2 + 3*n - 6) for k = 3 and n >= 3.

In general: For 1 <= k <= n with (n,k) <> 1,

T(n,k) = 2^(n-2)*Sum_{i=0..k-1} c(k,i)*(n-k+1)^(k-1-i)/(k-1)! and

T(n,k) = 2^(n-2)*Sum_{j=0..k-1} c(k,k-1-j)*(n-k+1)^j/(k-1)!

with c(k,i) being specific coefficients. Below are the first values for c(k,i):

1;

1, 1;

1, 7, 4;

1, 18, 77, 36;

1, 34, 359, 1238, 528,

1, 55, 1065, 8705, 26654, 10800;

... [Formula corrected by Petros Hadjicostas, Jul 15 2020]

The diagonal of this triangle for c(k,i) divided by (k-1)! (except for the first term) is equal to the Shroeder number sequence A006318(k-1).

From Petros Hadjicostas and Michel Marcus, Jul 15 2020: (Start)

T(n,1) = 2^(n-2) for n >= 2; T(n,k) = 2*(T(n,k-1) + T(n-1,k) - T(n-1,k-1)) for n > k >= 2; T(n,n) = 2*T(n,n-1) for n = k >= 2; and T(n,k) = 0 for 1 <= n < k. [Robeva and Sun (2020)] (They do not specify T(1,1) explicitly since they do not care about subdivisions of a degenerate polygon with only one side.)

T(n,k) = (2^(n-2)/(k-1)!) * P_k(n) = (2^(n-2)/(k-1)!) * Sum_{j=1..k} A336245(k,j)*n^(k-j) for n >= k >= 1 with (n,k) <> (1,1), where P_k(n) is some polynomial with integer coefficients of degree k-1. [Robeva and Sun (2020)]

A336245(k,j) = Sum_{s=0..j-1} c(k,s) * binomial(k-1-s, k-j) * (1-k)^(j-1-s) for 1 <= j <= k, in terms of the above coefficients c(k,i).

So c(k,s) = Sum_{j=1..s+1} A336245(k,j) * binomial(k-j, k-s-1) * (k-1)^(s+1-j) for k >= 1 and 0 <= s <= k-1, obtained by inverting the binomial transform.

Bivariate o.g.f.: x*y*(1 - x)*(1 - 2*y*g(2*x*y))/(1 - 2*x - 2*y + 2*x*y), where g(w) = 2/(1 + w + sqrt(1 - 6*w + w^2)) = g.f. of A001003.

Letting y = 1 in the above joint o.g.f., we get the o.g.f. of the row sums: x*(1-x)*(2*g(2*x) - 1). It can then be easily proved that

EXAMPLE

Triangle (with rows n >= 1 and columns k = 1..n) begins:

1;

1, 2;

2, 6, 12;

4, 16, 44, 88;

8, 40, 136, 360, 720;

16, 96, 384, 1216, 3152, 6304;

32, 224, 1024, 3712, 11296, 28896, 57792;

64, 512, 2624, 10624, 36416, 108032, 273856, 547712;

...

Example: T(4,3) = 44 = 1 + 1 + 2 + 2 + 6 + 12 + 4 + 16.

From Petros Hadjicostas, Jul 15 2020: (Start)

Consider the following 2-row grid with n = 3 points at the top and k = 2 points at the bottom:

A B C

*--*--*

| /

| /

*--*

D E

The sets of the dividing internal lines of the T(3,2) = 6 bimonotone subdivisions of the above 2-row grid are as follows: { }, {DC}, {DB}, {EB}, {DB, DC}, and {DB, EB}. We exclude subdivisions {EA} and {EA, EB} because they have at least one dividing line with a negative slope. (End)

PROG

(PARI) lista(nn) = {my(T=matrix(nn, nn)); T[1, 1] = 1; for (n=2, nn, for (k=1, n, T[n, k] = sum(i=1, n, sum(j=1, k, if ((i!=n) || (j!=k), T[i, j]))); ); ); vector(nn, k, vector(k, i, T[k, i])); } \\ Michel Marcus, Mar 18 2020

CROSSREFS

KEYWORD

nonn,tabl

AUTHOR

Andrea Raffetti, Jul 13 2011

EXTENSIONS

Offset changed by Andrew Howroyd, Dec 31 2017

Name edited by Petros Hadjicostas, Jul 15 2020

STATUS

approved