Recall that Euler's totient function $\varphi(n)$ is defined as the number of natural numbers $<n$ which are coprime to $n$. The Dirichlet series generating function for $\varphi(n)$ is $$ \sum_{n\geqslant1}\frac{\varphi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}. $$ The Lambert series is $$ \sum_{n\geqslant1}\frac{\varphi(n)q^n}{1-q^n}=\frac q{(1-q)^2}. $$ For the "ordinary" generating function the only identities I know involve the Möbius function, $$ \sum_{n\geqslant1}\varphi(n)x^n=\sum_{n\geqslant1}\frac{\mu(n)x^n}{(1-x^n)^2}=x\frac d{dx}\left(x\frac d{dx}\log\prod_{n\geqslant1}(1-x^n)^{-\frac{\mu(n)}{n^2}}\right). $$
Now what I need is a similar expression of any such kind for the generating function of $$ \varphi_2(n):=\#\left\{(a,b)\mid0<a,b<n<a+b,\gcd(a,b)=1\right\}. $$ The sequence of the $\varphi_2(n)$ for small $n$ goes like $$ 0,0,0,1,1,4,3,7,8,12,11,19,17,26,28,32,32,45,42,56,58,65,64,82,80,94,97,109,107,131, ... $$ It is not on OEIS
You have a square $[n-1]^2$ cut by the diagonal $x+y=n$.
The number of points in the square is $2\sum_{k=1}^{n-1}\phi(k)$.
Remove the points on the diagonals $x+y=k$, which is $\sum_{k=1}^{n}\phi(k)$ Final sum is $$-\phi(n)+\sum_{k=1}^{n-1}\phi(k)$$
Another way to look at it. Compare $\phi(n+1)$ with $\phi(n)$. Include points of the form $(a,n)$ and $(n,a)$, and remove points of the form $(a,(n+1)-a)$. If $a$ and $(n+1)-a$ are coprime then so are $a$ and $(n+1)-a+a$. So $$\phi_2(n+1)=\phi_2(n)+2\phi(n)-\phi(n+1)$$