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Let $a_{n,k}=a_{n,k}(x,c)$ be the generating function $\sum_P w(P),$ where $P$ runs over all paths from $(0,0)$ to $(n,k)$ consisting of horizontal steps $(1,0)$, up-steps $(1,1)$ and down-steps $(1,-1)$, which never run below the $x-$ axis and where $w(P)$ is the product of all weights of the steps of $P$ where the weights of up-steps and down-steps are $1$ and the weight of the horizontal steps on height $l>0$ are $s_l=c$ and on height $0$ are $s_0=c+x.$
With other words $$a_{n,k}= a_{n-1,k-1}+s_k a_{n-1,k}+ a_{n-1,k+1}$$ with $a_{0,k}=[k=0]$ and $a_{n,k}=0$ for $k<0.$

Let $$D_{k,n}= D_{k,n}(x,c) =\det_{0\leq i,j \leq {n-1}}(a_{i+j,k}(x,c)).$$ For $x=0$ and arbitrary $c$ it is known that $D_{k,(k+1)n}(0,c)=(-1)^{n \binom{k+1}{2}}$ and $ D_{k,n}=0$ else.

Computations for $k>0$ give

$\left(D_{1,n}(x,c)\right)_{n\geq 0}=\left(1,0,-1,x, 1-x^2, -2x+x^3, -1+3x^2-x^4,\dots \right),$ $\left(D_{2,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,-1,0,x^2, 1-x^2, x^2-x^4,-3x^2+2x^4,\dots \right),$ $\left(D_{3,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,0,1,0,0,x^3, 1-x^2,0, -x^4+x^6,\dots \right).$

Here we get $$D_{1,n+2}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-j-1}\binom{n-j}{j} x^{n-2j}$$ and $$D_{2,n+3}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-j-1}b_{j,n-2j}x^{2j}$$ with $\sum_{n\geq 0}{b_{k,n}z^n}=\frac{1}{(1-z)^k (1-z^3)^{k+1}}.$

For $k>2$ I could not find closed formulas, but it seems that $$D_{k,n}(x,c)=\det H_{k,n}(x),$$ where $H_{k,n}(x)$ has a rather simple form: $H_{k,n}(x)$ $=(h_{i,j,k})_{0\leq i,j\leq {n-1}}$ with $h_{k-m+l,m+l,k}=1$ if $0\leq m\leq k$ and $h_{l,k+i+l,k}=x^i $ if $i \geq 1$ and $l\geq 0$ and $h_{i,j,k}=0$ else. For example $$H_{2,6}(x)= \left ( \begin{matrix} 0 & 0 & 1& x & x^2 & x^3 \\ 0 & 1 & 0 &1 & x & x^2\\1 & 0 & 1 &0 & 1&x \\0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0\\0& 0 & 0 &1 &0 & 1 \end{matrix} \right ).$$

Any idea how to prove this?

Let $a_{n,k}=a_{n,k}(x,c)$ be the generating function $\sum_P w(P),$ where $P$ runs over all paths from $(0,0)$ to $(n,k)$ consisting of horizontal steps $(1,0)$, up-steps $(1,1)$ and down-steps $(1,-1)$, which never run below the $x-$ axis and where $w(P)$ is the product of all weights of the steps of $P$ where the weights of up-steps and down-steps are $1$ and the weight of the horizontal steps on height $l>0$ are $s_l=c$ and on height $0$ are $s_0=c+x.$
With other words $$a_{n,k}= a_{n-1,k-1}+s_k a_{n-1,k}+ a_{n-1,k+1}$$ with $a_{0,k}=[k=0]$ and $a_{n,k}=0$ for $k<0.$

Let $$D_{k,n}= D_{k,n}(x,c) =\det_{0\leq i,j \leq {n-1}}(a_{i+j,k}(x,c)).$$ For $x=0$ and arbitrary $c$ it is known that $D_{k,(k+1)n}(0,c)=(-1)^{n \binom{k+1}{2}}$ and $ D_{k,n}=0$ else.

Computations for $k>0$ give

$\left(D_{1,n}(x,c)\right)_{n\geq 0}=\left(1,0,-1,x, 1-x^2, -2x+x^3, -1+3x^2-x^4,\dots \right),$ $\left(D_{2,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,-1,0,x^2, 1-x^2, x^2-x^4,-3x^2+2x^4,\dots \right),$ $\left(D_{3,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,0,1,0,0,x^3, 1-x^2,0, -x^4+x^6,\dots \right).$

Here we get $$D_{1,n+2}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-j-1}\binom{n-j}{j} x^{n-2j}$$ and $$D_{2,n+3}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-j-1}b_{j,n-2j}x^{2j}$$ with $\sum_{n\geq 0}{b_{k,n}z^n}=\frac{1}{(1-z)^k (1-z^3)^{k+1}}.$

For $k>2$ I could not find closed formulas, but it seems that $$D_{k,n}(x,c)=d_{k,n}(x):=\det H_{k,n}(x),$$ where $H_{k,n}(x)$ has a rather simple form: $H_{k,n}(x)$ $=(h_{i,j,k})_{0\leq i,j\leq {n-1}}$ with $h_{k-m+l,m+l,k}=1$ if $0\leq m\leq k$ and $h_{l,k+i+l,k}=x^i $ if $i \geq 1$ and $l\geq 0$ and $h_{i,j,k}=0$ else. For example $$H_{2,6}(x)= \left ( \begin{matrix} 0 & 0 & 1& x & x^2 & x^3 \\ 0 & 1 & 0 &1 & x & x^2\\1 & 0 & 1 &0 & 1&x \\0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0\\0& 0 & 0 &1 &0 & 1 \end{matrix} \right ).$$

Computations also suggest that for $n \geq {2k+2}$ $d_{k,n}(x) d_{k,n-2k-2}(x)-x^2 d_{k,n-1}(x) d_{k,n-2k-1}(x)=(1-x^2)d_{k,n-k-1}^2.$

Any idea how to prove these conjectures?

Let $a_{n,k}=a_{n,k}(x,c)$ be the generating function $\sum_P w(P),$ where $P$ runs over all paths from $(0,0)$ to $(n,k)$ consisting of horizontal steps $(1,0)$, up-steps $(1,1)$ and down-steps $(1,-1)$, which never run below the $x-$ axis and where $w(P)$ is the product of all weights of the steps of $P$ where the weights of up-steps and down-steps are $1$ and the weight of the horizontal steps on height $l>0$ are $s_l=c$ and on height $0$ are $s_0=c+x.$
With other words $$a_{n,k}= a_{n-1,k-1}+s_k a_{n-1,k}+ a_{n-1,k+1}$$ with $a_{0,k}=[k=0]$ and $a_{n,k}=0$ for $k<0.$

Let $$D_{k,n}= D_{k,n}(x,c) =\det_{0\leq i,j \leq {n-1}}(a_{i+j,k}(x,c)).$$ For $x=0$ and arbitrary $c$ it is known that $D_{k,(k+1)n}(0,c)=(-1)^{n \binom{k+1}{2}}$ and $ D_{k,n}=0$ else.

Computations for $k>0$ give

$\left(D_{1,n}(x,c)\right)_{n\geq 0}=\left(1,0,-1,x, 1-x^2, -2x+x^3, -1+3x^2-x^4,\dots \right),$ $\left(D_{2,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,-1,0,x^2, 1-x^2, x^2-x^4,-3x^2+2x^4,\dots \right),$ $\left(D_{3,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,0,1,0,0,x^3, 1-x^2,0, -x^4+x^6,\dots \right).$

Here we get $$D_{1,n+2}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-k-1}\binom{n-k}{k} x^{n-2k}$$ and $$D_{2,n+3}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-k-1}b_{k,n-2k}x^{2k}$$ with $\sum_{n\geq 0}{b_{k,n}z^n}=\frac{1}{(1-z)^k (1-z^3)^{k+1}}.$

For $k>2$ I could not find closed formulas, but it seems that $$D_{k,n}(x,c)=\det H_{k,n}(x),$$ where $H_{k,n}(x)$ has a rather simple form: $H_{k,n}(x)$ $=(h_{i,j,k})_{0\leq i,j\leq {n-1}}$ with $h_{k-m+l,m+l,k}=1$ if $0\leq m\leq k$ and $h_{l,k+i+l,k}=x^i $ if $i \geq 1$ and $l\geq 0$ and $h_{i,j,k}=0$ else. For example $$H_{2,6}(x)= \left ( \begin{matrix} 0 & 0 & 1& x & x^2 & x^3 \\ 0 & 1 & 0 &1 & x & x^2\\1 & 0 & 1 &0 & 1&x \\0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0\\0& 0 & 0 &1 &0 & 1 \end{matrix} \right ).$$

Any idea how to prove this?

Let $a_{n,k}=a_{n,k}(x,c)$ be the generating function $\sum_P w(P),$ where $P$ runs over all paths from $(0,0)$ to $(n,k)$ consisting of horizontal steps $(1,0)$, up-steps $(1,1)$ and down-steps $(1,-1)$, which never run below the $x-$ axis and where $w(P)$ is the product of all weights of the steps of $P$ where the weights of up-steps and down-steps are $1$ and the weight of the horizontal steps on height $l>0$ are $s_l=c$ and on height $0$ are $s_0=c+x.$
With other words $$a_{n,k}= a_{n-1,k-1}+s_k a_{n-1,k}+ a_{n-1,k+1}$$ with $a_{0,k}=[k=0]$ and $a_{n,k}=0$ for $k<0.$

Let $$D_{k,n}= D_{k,n}(x,c) =\det_{0\leq i,j \leq {n-1}}(a_{i+j,k}(x,c)).$$ For $x=0$ and arbitrary $c$ it is known that $D_{k,(k+1)n}(0,c)=(-1)^{n \binom{k+1}{2}}$ and $ D_{k,n}=0$ else.

Computations for $k>0$ give

$\left(D_{1,n}(x,c)\right)_{n\geq 0}=\left(1,0,-1,x, 1-x^2, -2x+x^3, -1+3x^2-x^4,\dots \right),$ $\left(D_{2,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,-1,0,x^2, 1-x^2, x^2-x^4,-3x^2+2x^4,\dots \right),$ $\left(D_{3,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,0,1,0,0,x^3, 1-x^2,0, -x^4+x^6,\dots \right).$

Here we get $$D_{1,n+2}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-j-1}\binom{n-j}{j} x^{n-2j}$$ and $$D_{2,n+3}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-j-1}b_{j,n-2j}x^{2j}$$ with $\sum_{n\geq 0}{b_{k,n}z^n}=\frac{1}{(1-z)^k (1-z^3)^{k+1}}.$

For $k>2$ I could not find closed formulas, but it seems that $$D_{k,n}(x,c)=\det H_{k,n}(x),$$ where $H_{k,n}(x)$ has a rather simple form: $H_{k,n}(x)$ $=(h_{i,j,k})_{0\leq i,j\leq {n-1}}$ with $h_{k-m+l,m+l,k}=1$ if $0\leq m\leq k$ and $h_{l,k+i+l,k}=x^i $ if $i \geq 1$ and $l\geq 0$ and $h_{i,j,k}=0$ else. For example $$H_{2,6}(x)= \left ( \begin{matrix} 0 & 0 & 1& x & x^2 & x^3 \\ 0 & 1 & 0 &1 & x & x^2\\1 & 0 & 1 &0 & 1&x \\0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0\\0& 0 & 0 &1 &0 & 1 \end{matrix} \right ).$$

Any idea how to prove this?

Let $a_{n,k}=a_{n,k}(x,c)$ be the generating function $\sum_P w(P),$ where $P$ runs over all paths from $(0,0)$ to $(n,k)$ consisting of horizontal steps $(1,0)$, up-steps $(1,1)$ and down-steps $(1,-1)$, which never run below the $x-$ axis and where $w(P)$ is the product of all weights of the steps of $P$ where the weights of up-steps and down-steps are $1$ and the weight of the horizontal steps on height $l>0$ are $s_l=c$ and on height $0$ are $s_0=c+x.$
With other words $$a_{n,k}= a_{n-1,k-1}+s_k a_{n-1,k}+ a_{n-1,k+1}$$ with $a_{0,k}=[k=0]$ and $a_{n,k}=0$ for $k<0.$

Let $$D_{k,n}= D_{k,n}(x,c) =\det_{0\leq i,j \leq {n-1}}(a_{i+j,k}(x,c)).$$ For $x=0$ and arbitrary $c$ it is known that $D_{k,(k+1)n}(0,c)=(-1)^{n \binom{k+1}{2}}$ and $ D_{k,n}=0$ else.

Computations for $k>0$ give

$\left(D_{1,n}(x,c)\right)_{n\geq 0}=\left(1,0,-1,x, 1-x^2, -2x+x^3, -1+3x^2-x^4,\dots \right),$ $\left(D_{2,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,-1,0,x^2, 1-x^2, x^2-x^4,-3x^2+2x^4,\dots \right),$ $\left(D_{3,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,0,1,0,0,x^3, 1-x^2,0, -x^4+x^6,\dots \right).$

I could not find closed formulas, but it seems that $$D_{k,n}(x,c)=\det H_{k,n}(x),$$ where $H_{k,n}(x)$ has a rather simple form: $H_{k,n}(x)$ $=(h_{i,j,k})_{0\leq i,j\leq {n-1}}$ with $h_{k-m+l,m+l,k}=1$ if $0\leq m\leq k$ and $h_{l,k+i+l,k}=x^i $ if $i \geq 1$ and $l\geq 0$ and $h_{i,j,k}=0$ else. For example $$H_{2,6}(x)= \left ( \begin{matrix} 0 & 0 & 1& x & x^2 & x^3 \\ 0 & 1 & 0 &1 & x & x^2\\1 & 0 & 1 &0 & 1&x \\0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0\\0& 0 & 0 &1 &0 & 1 \end{matrix} \right ).$$

Any idea how to prove this?

Let $a_{n,k}=a_{n,k}(x,c)$ be the generating function $\sum_P w(P),$ where $P$ runs over all paths from $(0,0)$ to $(n,k)$ consisting of horizontal steps $(1,0)$, up-steps $(1,1)$ and down-steps $(1,-1)$, which never run below the $x-$ axis and where $w(P)$ is the product of all weights of the steps of $P$ where the weights of up-steps and down-steps are $1$ and the weight of the horizontal steps on height $l>0$ are $s_l=c$ and on height $0$ are $s_0=c+x.$
With other words $$a_{n,k}= a_{n-1,k-1}+s_k a_{n-1,k}+ a_{n-1,k+1}$$ with $a_{0,k}=[k=0]$ and $a_{n,k}=0$ for $k<0.$

Let $$D_{k,n}= D_{k,n}(x,c) =\det_{0\leq i,j \leq {n-1}}(a_{i+j,k}(x,c)).$$ For $x=0$ and arbitrary $c$ it is known that $D_{k,(k+1)n}(0,c)=(-1)^{n \binom{k+1}{2}}$ and $ D_{k,n}=0$ else.

Computations for $k>0$ give

$\left(D_{1,n}(x,c)\right)_{n\geq 0}=\left(1,0,-1,x, 1-x^2, -2x+x^3, -1+3x^2-x^4,\dots \right),$ $\left(D_{2,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,-1,0,x^2, 1-x^2, x^2-x^4,-3x^2+2x^4,\dots \right),$ $\left(D_{3,n}(x,c)\right)_{n\geq 0}=\left(1,0,0,0,1,0,0,x^3, 1-x^2,0, -x^4+x^6,\dots \right).$

Here we get $$D_{1,n+2}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-k-1}\binom{n-k}{k} x^{n-2k}$$ and $$D_{2,n+3}(x,c)=\sum_{0\leq 2j\leq n}(-1)^{n-k-1}b_{k,n-2k}x^{2k}$$ with $\sum_{n\geq 0}{b_{k,n}z^n}=\frac{1}{(1-z)^k (1-z^3)^{k+1}}.$

For $k>2$ I could not find closed formulas, but it seems that $$D_{k,n}(x,c)=\det H_{k,n}(x),$$ where $H_{k,n}(x)$ has a rather simple form: $H_{k,n}(x)$ $=(h_{i,j,k})_{0\leq i,j\leq {n-1}}$ with $h_{k-m+l,m+l,k}=1$ if $0\leq m\leq k$ and $h_{l,k+i+l,k}=x^i $ if $i \geq 1$ and $l\geq 0$ and $h_{i,j,k}=0$ else. For example $$H_{2,6}(x)= \left ( \begin{matrix} 0 & 0 & 1& x & x^2 & x^3 \\ 0 & 1 & 0 &1 & x & x^2\\1 & 0 & 1 &0 & 1&x \\0 & 1 & 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0\\0& 0 & 0 &1 &0 & 1 \end{matrix} \right ).$$

Any idea how to prove this?