Finding general term of $\{x_n\}$ given $x_{m+n} = x_m + x_n + m\cdot n, \; \forall m, n \in \mathbb N$ 
There exists a sequence $\{x_n\}$ members of which satisfy the following equation:
$$x_{m+n} = x_m + x_n + m\cdot n,\; \forall m, n \in \mathbb N, \; x_1 = a
$$
  Find the general term of the sequence.

To solve this I started from the case when $m = n = 1$. That gives:
$$
x_2 = x_1 + x_1 +  = 2x_1 +1
$$
I want to look at $x_3, x_4, ..., x_n$ and find some pattern. So for $m = 1$ and $n = 2$ we obtain:
$$
x_3 = x_1 + x_2 + 2 = x_1 + 2x_1 + 1 = 3x_1 + 3
$$
Continuing that process we obtain:
$$
x_1 = \color{green}1x_1 + \color{blue}0  \\
x_2 = \color{green}2x_1 + \color{blue}1 \\
x_3 = \color{green}3x_1 + \color{blue}3 \\
x_4 = \color{green}4x_1 + \color{blue}8 \\
x_5 = \color{green}5x_1 + \color{blue}{10} \\
x_6 = \color{green}6x_1 + \color{blue}{15}
$$
And so on. The green values form a sequence of natural numbers. For the blue ones i referred to OEIS and the sequence is in the form ${1 \over 2} (n-1)$. Based on that the general term is:
$$
x_n = n(a + {1\over 2}(n-1))
$$
I've also looked at a similar sequence which satisfies the following:
$$
x_{m+n} = x_m + x_n + m + n,\; \forall m, n \in \mathbb N
$$
And again I was able to search for the sequence in OEIS but could not deduce it myself without a lookup. Also it is not going to always work since one may imagine sequences which match for the first $n$ terms and starting from $n+1$ become different.
My question is:

I'm interested whether there exists a common approach to find the generating function for a given sequence (of numbers). Or perhaps the problem above may be solved in a different way which would give the general term formula without "guessing" its form.

 A: Let $f(n) = x_n$ then for $m=1$ we have $$f(n+1) = f(n)+n+a$$
Then by telescoping we get $$ f(n) = na+{n(n-1)\over 2}$$
A: An alternative method is given here (for the equation $x_{m+n}=x_m+x_n+mn$).  Let $$f(k):=x_k-\frac{1}{2}\,k^2\text{ for every }k\in\mathbb{Z}_{>0}\,.$$
Then, $f$ satisfies Cauchy's functional equation:
$$f(m+n)=f(m)+f(n)\text{ for all }m,n\in\mathbb{Z}_{>0}\,.$$
It is well known that the solution is given by $f(k)=\lambda\,k$, where $\lambda$ is a fixed constant.  Therefore,
$$x_k=\lambda\,k+\frac{1}{2}\,k^2\text{ for }k=1,2,\ldots\,.$$
As $x_1=a$, we get $\lambda+\dfrac12=a$, whence
$$x_k=\left(a-\frac{1}{2}\right)\,k+\frac{1}{2}\,k^2=a\,k+\frac{k(k-1)}{2}\text{ for each }k=1,2,3,\ldots\,.$$

For the equation $x_{m+n}=x_m+x_n+m+n$, we note that $x_2=2x_1+2$ and so
$$x_3=x_1+x_2+3=x_1+(2x_1+2)+3=3x_1+5\,.$$
Thus,
$$x_4=2x_2+4=4x_1+8$$
but
$$x_4=x_1+x_3+4=x_1+(3x_1+5)+4=4x_1+9\,,$$
that is, $4x_1+8=x_4=4x_1+9$, which is a contradiction.  Therefore, such a sequence does not exist.


More generally, let $g:\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}\to\mathbb{C}$ be an arbitrary function.  Then, there exists a sequence $\left(x_k\right)_{k\in\mathbb{Z}_{>0}}$ of complex numbers such that
$$x_{m+n}=x_m+x_n-g(m,n)\text{ for every }m,n\in\mathbb{Z}_{>0}$$
if and only if
$$g(m,n)=\sum_{s=m}^{m+n-1}\,g(s,1)-\sum_{s=1}^{n-1}\,g(s,1)\text{ for every }m,n\in\mathbb{Z}_{>0}\,.\tag{*}$$

First, we shall prove the converse.  Define
$$h(k):=\sum_{s=1}^{k-1}\,g(s,1)\text{ for }k=1,2,3,\ldots\,.$$
From (*), it follows immediately that
$$h(m+n)=h(m)+h(n)+g(m,n)\text{ for each }m,n\in\mathbb{Z}_{>0}\,.$$
Thus, the function $f:\mathbb{Z}_{>0}\to\mathbb{C}$ defined by $f(k)=x_k-h(k)$ for every $k=1,2,3\ldots$ satisfies Cauchy's functional equation.  That is, $f(k)=\lambda\,k$ for some fixed constant $\lambda\in\mathbb{C}$.  Hence,
$$x_k=\lambda\,k+h(k)=\lambda\,k+\sum_{s=1}^{k-1}\,g(s,1)\text{ for any }k=1,2,3,\ldots\,.$$
The value $x_1$ determines the whole sequence (noting that $\lambda=x_1$).
For the direct implication, we note that
$$x_{s+1}=x_s+x_1-g(s,1)\text{ for every }s=1,2,3,\ldots\,.$$
That is,
$$x_k-k\,x_1=\sum_{s=1}^{k-1}\,\big(x_{s+1}-x_s-x_1\big)=\sum_{s=1}^{k-1}\,g(s,1)\text{ for all }k=1,2,3,\ldots\,.$$
Thus, for each $k\in\mathbb{Z}_{>0}$, we obtain
$$x_k=k\,x_1+h(k)\,,\text{ where }h(k):=\sum_{s=1}^{k-1}\,g(s,1)\,.$$
Since $x_{m+n}=x_m+x_n+g(m,n)$, we finally obtain
$$h(m+n)=h(m)+h(n)+g(m,n)\text{ for all }m,n\in\mathbb{Z}_{>0}\,,$$
which is equivalent to (*).
Remark:  For a sequence $\left(x_k\right)_{k\in\mathbb{Z}_{>0}}$ to exist, it is necessary that $g$ is symmetric in its two arguments.  It is important to note that, if $g$ satisfies (*), then $g$ is a symmetric function.

We can apply the general result to the OP's problem.  For the equation $x_{m+n}=x_m+x_n+mn$, we have $g(m,n)=mn$.  Therefore,
$$\sum_{s=m}^{m+n-1}\,g(s,1)-\sum_{s=1}^{n-1}\,g(s,1)=n\left(m+\frac{n-1}{2}\right)-\frac{n(n-1)}{2}=mn=g(m,n)\,.$$
Therefore, there exists such a sequence, and it is given by
$$x_k=k\,x_1+\sum_{s=1}^{k-1}\,g(s,1)=a\,k+\frac{k(k-1)}{2}\,.$$
For the equation $x_{m+n}=x_m+x_n+m+n$, we obtain $g(m,n)=m+n$.  Note that
$$\sum_{s=m}^{m+n-1}\,g(s,1)-\sum_{s=1}^{n-1}\,g(s,1)=n\left(m+\frac{n+1}{2}\right)-\frac{(n-1)(n+2)}{2}=mn+1\,.$$
Thus, (*) is not satisfied, and such a sequence cannot exist.
