The precise general expression for the sequence generated by my code

The code can be translated thus

$$\sum_{j=1}^N\sum_{i=1}^{\max(2j-1,N)}\left(2j-i\right). $$

Now let $N = 2m$ then $2j-1<N$ is equivalent to $j\le m$. So we split the sum:

$$\begin{align} &\sum_{j=1}^m\sum_{i=1}^{2j-1}(2j-i) + \sum_{j=m+1}^{2m}\sum_{i=1}^{N}(2j-i) \\ &= \sum_{j=1}^m \left((2j)(2j-1)-\frac{(2j-1)(2j)}{2}\right) + \sum_{j=m+1}^{2m} \left(2jN-\frac{N(N+1)}{2}\right) \\ &= \sum_{j=1}^m\left( 2j^2 - j \right) + \left(2\cdot 2m\left(\sum_{j=1}^{2m}j-\sum_{j=1}^m j\right) - \frac{2m(2m+1)}{2}\cdot m\right)\\ &= 2 \left(\frac{m(m+1)(2m+1)}{6} \right) - \frac{m(m+1)}{2} + \left(2m(4m^2+2m-(m^2+m))-m(2m+1)(m)\right)\\ &= m(m+1)\left(\frac{2m+1}{3}-\frac12\right) + \left(2m(3m^2+m)-m^2(2m+1)\right)\\ &= \frac16(m)(m+1)(4m-1)+m^2(4m+1)\\ &=\frac16 m(28m^2+9m-1). \end{align}$$

Now you should do the case of $N=2m+1$, and noting that 2j-1<N is also equivalent to $j\le m$. Then, the first sum is the same as above.

Correcting the typos in my comment:

Your sum is $$\sum_{i=1}^n \sum_{j=1}^n \sum_{k=i}^{2j-1} 1 = \sum_{j=1}^n \sum_{i=1}^{\max(2j, n)} (2j-i)$$ Splitting the $j$ sum into parts where $2j\leq n$ and $2j>n$, using the formulas for $\sum j$ and $\sum j^2$, and conditioning on whether $n$ is odd or even, the sum is $(28m^3 + 9m^2 - m)/6$ if $n=2m$ (values 6, 43, 139, ...) and $(m + 1) (28 m^2 + 23 m + 6)/6$ if $n=2m+1$ (values 1, 19, 82, ...).

For example, if $n=2m+1$ then the sum is $\sum_{j=1}^m \sum_{i=1}^{2j} (2j-i) + \sum_{j=m+1}^{2m+1} \sum_{i=1}^n (2j-i)$. The sum from $i=1$ to $2j$ of $(2j-i)$ is $j(2j-1)$ and the sum from $i=1$ to $n$ of $(2j-i)$ is $2jn-n(n+1)/2$, so we get

$$ \sum_{j=1}^m j(2j-1) + \sum_{j=m+1}^{2m+1}(2jn-n(n+1)/2)$$

Using the sum-of-squares formula, this is

$$ m(m+1)(2m+1)/3 - m(m+1)/2 + 2(2m+1)(m+1)(3m+2)/2 - (m+1)^2(2m+1) $$

which simplifies to $(m+1)(28m^2+23m+6)/6$.