I just stumbled on this interesting function $$(n* \sum(n+1))/3$$ but I can't find any reference to it online for tetrahedra numbers.
I also divided $$(n* \sum(n+1)) $$ by a known nth tetrahedra formula $$(n * (n + 1) * (n + 2)) / 6 $$ which gave me 3 for the first 20K values.
Is it correct to assume that $$(n* \sum(n+1))/3 $$ is a valid formula for nth tetrahedra numbers
Link to Github(After line 141): https://github.com/JJpysquared/RIF/blob/master/Escanor
Yes, assuming you mean $$\frac{n\sum_{i=1}^{n+1}i}3.$$ The reason for this, is, of course, that the summation is just $$\frac{(n+1)(n+2)}2,$$ and thus the whole expression trivially evaluates to $$\frac{n(n+1)(n+2)}6.$$