Here is Vassil's solution:

Let $f(n)$ denote the sum of the first $n$ terms of the sequence

${\displaystyle 0,1,1,2,2,3,3,\dots ,p,p,p+1,p+1,\dots .}$

First I tried with several numbers. Let $n=15$. Then $f(15)=2*(1+2+3+4+5+6+7)=7*8$ where the sequence is: 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7

Let $n=14$. Then $f(14)=2*(1+2+3+4+5+6+7)-7=7*7$ where the sequence is: 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7

Let $n=17$. Then $f(17)=2*(1+2+3+4+5+6+7+8)=8*9$ where the sequence is: 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8

Let $n=16$. Then $f(16)=2*(1+2+3+4+5+6+7+8)-8=8*8$ where the sequence is: 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8

I noticed that the formula for $f(n)$ depends on whether $n$ is odd or even.

Case I - $n$ is odd, i.e. $n=2k+1$

Then

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(n)}& =2(1+2+...+k)\\ \multicolumn{1}{c}{}& =2k(k+1)/2\\ \multicolumn{1}{c}{}& =\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right).\end{array}}$

Case II - $n$ is even, i.e. $n=2k$

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(n)}& =2(1+2+...+k)-k\\ \multicolumn{1}{c}{}& =k^2+k-k\\ \multicolumn{1}{c}{}& =k^2\\ \multicolumn{1}{c}{}& ={\left(\frac{n}{2}\right)}^2.\end{array}}$

Now we have to calculate $f(a+b)-f(a-b)$

There are two cases. In the first case, when one of $a$ and $b$ is even and the other is odd, then $(a+b)$ and $(a-b)$ are both odd. Otherwise $(a+b)$ and $(a-b)$ are both even.

Case I $(a+b)$ and $(a-b)$ both odd.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(a+b)-f(a-b)}& =\frac{(a+b{)}^2-1}{4}-\frac{(a-b{)}^2-1}{4}\\ \multicolumn{1}{c}{}& =\mathrm{ab}.\end{array}}$

Case II $(a+b)$ and $(a-b)$ both even.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(a+b)-f(a-b)}& =\frac{(a+b{)}^2}{4}-\frac{(a-b{)}^2}{4}\\ \multicolumn{1}{c}{}& =\mathrm{ab}.\end{array}}$