Number of solution of Frobenius equation Oke I am trying to find all presentable of a number $n$ as sum $ax+(a+1)y$ where $a=0,1,\ldots$ and $x,y\geq0$ are integers. 
I find that
$5=1+1+1+1+1=1+1+1+2=1+2+2=2+3=5$ so we have $5$ ways.
$7=1+1+1+1+1+1+1=1+1+1+1+1+2=1+1+1+2+2=1+2+2+2=2+2+3=3+4$ so we have $7$ ways.
I suppose that if $p\geq3$ is a prime number then there are exactly $p$ ways. However, I found that
$11=1+1+1+1+1+1+1+1+1+1+1=1+1+1+1+1+1+1+1+1+2=1+1+1+1+1+1+1+2+2=1+1+1+1+1+2+2+2=1+1+1+2+2+2+2=1+2+2+2+2+2$
and
$=2+2+2+2+3=2+3+3+3=3+4+4 =11$ so there are only $10$ ways. 
My question are that: What is the formula for the number ways to preprent a prime $p$ to sum of $x$ numbers $a$ and $y$ numbers $a+1$ where $a,x,y$ are non-negative integers?

This should hold for any $n$, not just $n$ prime. For the induction step from $n-1$ to $n$, note that a generic sum is
$$n-1 = ax + (a+1)y,$$
or
$$n = ax + (a+1)y +1 = a(x-1) + (a+1)(y+1),$$
assuming $x$ is not zero. So in this case, a sum for $n-1$ corresponds to one for $n$. If x is zero, then use
$$n-1= ay,$$
or 
$$n = ay+1= a(y-1) + (a+1),$$
and we again relate a sum in $n-1$ to one in $n$. 
Additionally, you have a new sum of $n$ as $n$ 1s, so the induction step adds 1 breakdown over that for $n-1$.
This needs fleshing out (inversing the step, etc), of course. 
