I have the recurrence relation $g(n,k)=g(n-2,k-1) + g(n-1,k)$ for all $k\geq1$ with the boundary conditions $g(n,k)=0$ if $n<2k-1$ and $g(2k-1,k)=1$
What I'm trying to do is define a new function by the equation $ h(p,q) = g(n,k)$ where $n=p+q-1$ and $k=q$ and show that $ h(p,q)$ satisfies the binomial recurrence relation and its boundary conditions and hence deduce that $g(n,k)=\binom{n-k+1}{k}$
My workings are:
We have the binomial recurrence relation $\binom{p}{q}=\binom{p-1}{q-1}+\binom{p-1}{q}$
where$\begin{align*} \\&\binom{p}{q}=0\text{ when }q>p \\&\binom{p}{q}=1\text{ when }p=q \end{align*}$ Now if $\begin{align*} \\&p=n-k+1 \\&q=k \end{align*}$
Then $\begin{align*} \\&\binom{n-k+1}{k}=0\mbox{ when }n<2k-1\text{ (the first boundary condition)}\end{align*}$ and $\begin{align*} \\&\binom{n-k+1}{k}=1\text{ when }n=2k-1 \text{ (the second boundary condition)}\end{align*}$
Also $\begin{align*}\\&\binom{n-k+1}{k}=\binom{n-k}{k-1}+\binom{n-k}{k}\end{align*}\tag{a}$
Now from the recurrence relation $g(n,k)$ we have (Is this true???): $\binom{n}{k}=\binom{n-2}{k-1}+\binom{n-1}{k}$ Which therefore equals: $ \binom{n-k+1}{k}=\binom{n-k-1}{k-1}+\binom{n-k}{k}\tag{b} $ Setting $(a)$ equal to $(b)$ we now get: $\begin{align*}\binom{n-k}{k-1}+\binom{n-k}{k}&=\binom{n-k-1}{k-1}+\binom{n-k}{k}\\ \binom{n-k}{k-1}&=\binom{n-k-1}{k-1}\end{align*}$
However, I am not sure where to go from here. I'm rather dubious as to the veracity of this identity as I have been unable to prove that it is true. Am I on the right track? If so how do I go about proving this identity? If I'm off beam any ideas where I went wrong?
Thanks in advance.