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;
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).
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
Sum_{k=1..n} T(n,k) = 2^n*
A001003(n-1) - 2^(n-1)*
A001003(n-2) for n >= 3. (End)
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