On the number of solutions of a type of diophantine equation For $n \in \mathbb N$, define $f(n):=|\{(a,b,c)\in \mathbb N^3 : n=2a+b+c , a\ne b , a\ne c , b<c\}|$ . How to show that there are infinitely many values of $n$ such that $f(n+1)<f(n)$ ? What can we say about lim inf and lim sup of the sequence $\{f(n+1)-f(n)\}$  ? ( like at least finite , positive etc. ? if it can be shown that $f(n+1)<f(n)$ for infinitely many positive integer $n$, then of-course $\lim \inf \{f(n+1)-f(n)\}<0$ )  
 A: I attempted to answer the question by trying to express $f(n+1)-f(n)$ as a the cardinal of some set but without success. notwithstanding all of that, I think there should be a simpler approach by exhibiting some combinatoric properties of the sets in the question. Anyway, let's kill a fly with a sledgehammer and compute $f(n)$ (maybe the given expression can be simplified ?):

Claim: We will prove that : $$f(n)=  \frac{\lfloor\frac{n+1}{2} \rfloor\left(\lfloor\frac{n+1}{2} \rfloor+1\right)}{2}-\lfloor \frac{n}{3} \rfloor -1+\epsilon(n)$$
  where $\epsilon(n)=1 $ if $n\pmod4=0$ and $0$ otherwise.

Now assuming this is proven (and hopefully without any mistake), we have : 
$$
\begin{align*}
f(2k+1)-f(2k)=k+1-\epsilon(2k)+\lfloor \frac{2k}3\rfloor -\frac{2k+2}3\rfloor \\
f(2k)-f(2k-1)=\epsilon(2k)+\lfloor \frac{2k-1}3\rfloor -\frac{2k}3\rfloor
\end{align*}
$$
From here you can deduce that $f(n)-f(n-1)$ can be as large as you want, and $f(6k)-f(6k-1)=\epsilon(6k)-1$ which is equal to $-1$ only when $n=12k+6$ and equal to $0$ when $n=12k$. You can also - using case analysis modulo $12$ - prove that the only numbers that don't appear in the sequence $f(n+1)-f(n)$ are those of the form $6m+5$.

Proof : 
Let's define $F(n)=\{(a,b,c)\in \mathbb N^3 : n=2a+b+c , a\ne b , a\ne c , b<c\}$ 
so that we have $f(n)=|F(n)|$. Now let's assume that $a$ is fixed, then we will have to compute the cardinal of the set defined as follows (for some all values of $n$) $G(n)=\{(a,b)\in \mathbb N^2 : n=a+b , a\ne b, a<b\}$ it's easy to see that :
$$g(n)=|G(n)|=\lfloor\frac{n+1}{2} \rfloor $$
Finally, define $H(n,a)=\{(b,c)\in G(n-2a): a\ne b , a\ne c\} $ so that we have the following equivalence :
$(a,b,c)\in F(n) \iff (b,c)\in H(n,a) $ .
This is just an alternative way of phrasing the same question, but from here we can actually compute the size of $H$ and then the size of $F$. As for $H$ :
$$ h(n,a)=|H(n,a)|=g(n-2a)-\epsilon(n,a)$$
where $\epsilon(n,a) =0$ except when $a$ appears in some tuple in $G(n)$ in which case $\epsilon(n,a) =1$, this is equivalent to $a\leq n-2a$ and $n\ne 4a$. Now it's time to compute the size of $F$ :
$$\begin{align*}
f(n) &= \sum_{0\leq 2a\leq n} h(n,a)\\
&=\sum_{0\leq 2a\leq n} g(n-2a) - \sum_{0\leq 2a\leq n} \epsilon(n,a)\\
&=\sum_{0\leq k\leq \lfloor\frac{n}{2}\rfloor} g(2k+p(n)) - \sum_{0\leq 2a\leq n,3a\leq n, 4a\ne n} 1 \tag{1}\\
&=\sum_{0\leq k\leq \lfloor\frac{n}{2}\rfloor} (k+p(n)) - \left( \lfloor \frac{n}{3} \rfloor +1-\epsilon(n) \right) \tag 2\\ \\
&= \frac{\lfloor\frac{n+1}{2} \rfloor\left(\lfloor\frac{n+1}{2} \rfloor+1\right)}{2}-\lfloor \frac{n}{3} \rfloor -1+\epsilon(n)
\end{align*}$$


*

*When $a$ verifies $0\leq 2a\leq n$ then $n-a$ runs over all the even numbers if $n$ is even, and over the odd the numbers if $n$ is odd  that are less than or equal to $n$, note that $p(n)=1$ if $n$ is odd and $p(n)=0$ otherwise.

*Evaluating the second sum is equivalent to counting the number of integers $a$ verifying $0\leq 2a\leq n,3a\leq n, 4a\ne n$ or equivalently $0\leq 3a\leq n, 4a\ne n $. Note that $\epsilon(n)=0 $ except when $n$ is divisible by $4$ for which $\epsilon(n)=1$.



