# Sum of spaced binomial coefficients as a Congruence

Suppose that I have a recurrence relation

$$\sum_{k=0}^n\binom{mn}{mk}A(mk)=0 \Rightarrow A(mn)=-\sum_{k=0}^{n-1}\binom{mn}{mk}A(mk)$$

with conditions $A(0)=1$, and $A(n)=0$ if $n\not\equiv 0\pmod{m}$. Then using the recurrence relation, it is easy to get that for $n=1, A(m)=-1$. Then for $n=2$,

$$A(2m)=-\sum_{k=0}^1\binom{2m}{km}A(km)=-\binom{2m}{0}A(0)-\binom{2m}{m}A(m)=-1+\binom{2m}{m}$$

By the Catalan number relation

$$\binom{2m}{m}=C(m)(m+1)$$

where the factor $C(m)$ is the $m$-th Catalan Number, it is seen that $A(2m)\equiv -1\pmod{m+1}$

Using these facts, I'd like to try and show that

$$A(mn)\equiv -1{\pmod{m+1}}$$

I thought induction would help, so i assume for all $k\ge n, A(mk)\equiv -1\pmod{m+1}$, and though would mean that

$$A(m(n+1))=-\sum_{k=0}^n\binom{mn}{mk}A(mk)\equiv\sum_{k=0}^n\binom{mn}{mk}\pmod{m+1}$$

However, from here, I am stuck. Can anyone either point me in the right direction, or show me if there is a fatal flaw in my reasoning?

I'd like to try and show that $$A(mn)\equiv -1{\pmod{m+1}}$$
This does not hold for $(m,n)=(5,4)$.
We have $$\sum_{k=0}^1\binom{5\times 1}{5k}A(5k)=0\implies A(5)=-1$$ $$\sum_{k=0}^2\binom{5\times 2}{5k}A(5k)=0\implies A(10)=251$$ $$\sum_{k=0}^3\binom{5\times 3}{5k}A(5k)=0\implies A(15)=-750751$$ $$\sum_{k=0}^4\binom{5\times 4}{5k}A(5k)=0\implies A(20)=11593285251$$ and $$A(20)\equiv 3\pmod 6$$