Let $p(x) = (1+x+x^2)^d$ for $d\ge 2$ and call its coefficients $$ p(x) = a_0 + a_1x+ a_2x^2 + \dots + a_{2d} x^{2d}. $$ Let $T(d)$ be the infinite upper triangular and Toeplitz matrix defined as $$ T(d)=\begin{pmatrix} a_0 & a_1 & a_2 & \dots & a_{2d-1} & a_{2d} & 0 & 0 & \dots & &\\ 0 & a_0 & a_1 & a_2 & \dots & a_{2d-1} & a_{2d} & 0 & 0 & \dots &\\ 0 & 0 & a_0 & a_1 & a_2 & \dots & a_{2d-1} & a_{2d} & 0 & 0 & \dots\\ & & & \ddots & \ddots & \ddots & & \ddots & \ddots & &\\ \end{pmatrix} $$ for every $n\ge 1$ take the following submatrices: $$ A_{d,n,1} = [T(d)]_{i=1:n}^{j=1:n}, \quad A_{d,n,2} = [T(d)]_{i=1:n}^{j=2:n+1}, \quad A_{d,n,3} = [T(d)]_{i=1:n}^{j=3:n+2}, \quad \dots \quad A_{d,n,2d+1} = [T(d)]_{i=1:n}^{j=2d+1:n+2d}, $$ that are all the square submatrices you can extract from the first $n$ rows with consecutive columns and with non-zero diagonal.
Claim: $A_{d,n,k}$ is singular only for $k=2$ or $k=2d$ and the couples $$(d,n) = (4,3), (10, 8), (12,5)$$
We know a number of facts already, that are
- $det(A_{d,n,1}) = det(A_{d,n,2d+1}) = 1$
- $A_{2d,n,2d+1}$ are positive definite
- $det(A_{d,n,k})$ diverges to $+\infty$ with $n$ if $k$ is odd and $2<k<2d$
The claim is related to Schur symmetric polynomials (since we are working with Toeplitz matrices) and also it would mean that
- If $c_{n,d}$ is the n-th coefficient of the formal expansion of $$\frac 1{(1+x+x^2)^d} $$ then $$ c_{n,d} = 0 \iff (d,n) = (4,3), (10, 8), (12,5)$$
- $$ \sum_{r=0}^n (-1)^k {d+r-1 \choose r}{r \choose n-r}=0 \iff (d,n) = (4,3), (10, 8), (12,5)$$
- $$\frac{\partial^n}{\partial x^n}\left[(1-x^2)^{\frac{n+d-1}2}\right]\Big|_{x=1/2} =0 \iff (d,n) = (4,3), (10, 8), (12,5) $$
- If $c_n^{(\alpha)}(x)$ are the Gegenbauer polynomials, then $$c_n^d(\frac 12) = 0 \iff (d,n) = (4,3), (10, 8), (12,5)$$
- If $P_n^{(\alpha,\beta)}(x)$ are the Jacobi polynomials, then $$P_n^{(-1/2,d-1/2)}(\frac 12) = 0 \iff (d,n) = (4,3), (10, 8), (12,5)$$
- If $ _2F_1(a,b,c,z)$ are the Hypergeometric functions, then $$ _2F_1(-n,d+n,\frac 12,\frac 14) = 0 \iff (d,n) = (10, 4)$$ $$_2F_1(-n,d+n+1,\frac 32,\frac 14) = 0 \iff (d,n) = (4,1), (12,2) $$
- Let $b_n^{(d)}$ be the recurrence sequence defined by $$b_0^{(d)} = 1, \quad b_1^{(d)}=d, \quad nb_n^{(d)} = (n+d-1)b_{n-1}^{(d)} - (n+2d-2)b_{n-2}^{(d)}.$$ Then $$b_n^{(d)} = 0 \iff (d,n) = (4,3), (10, 8), (12,5)$$
If you have any insight on any of the above cited problems, please tell.
An interesting problem, don't have much to add without spending more time, than that the generating function for your matrices $A_{d,n,k}=T_n(f_{d,n,k})$ are
$$ f_{d,n,k}(\theta)=e^{k\mathbf{i}\theta}\sum_{j=0}^{2d} {{d+1}\choose{j}}_2 e^{-j\mathbf{i}\theta} $$
where ${{d+1}\choose{j}}_2$ is the entries of the trinomial triangle. Where does this problem come from?