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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

JJ-Pysquared
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1 Answers1

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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 $$\frac{n(n+1)(n+2)}6.$$

hdighfan
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  • Thank you for the explanation. – JJ-Pysquared Jun 12 '20 at 17:15
  • I ended up finding it while trying to find functions that equals to 3 for all given inputs but when I finally found a function that contains 3 as a factor for all values I divided it by 3 and the result was the tetrahedra numbers which I've only heard of before but dividing it by the Tetrahedra formula allowed me to account for all other factors other than 3 and gave me the desired 3 for all values. – JJ-Pysquared Jun 12 '20 at 17:23