Which Day of Christmas is it?
Python, 27 bytes



lambda n:int((n*6)**.33359)


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A direct formula with some curve-fitting, same as the original one found by Level River St. 

The shifted equation i**3-i==n*6 is close to i**3==n*6 for large i. It solves to i=(n*6)**(1/3). Taking the floor the rounds down as needed, compensating for the off-by-one.

But, there are 6 inputs on boundaries where the error takes it below an integer it should be above. All of these can be fixed by slightly increasing the exponent without introducing further errors.



Python, 38 bytes



f=lambda n,i=1:i**3-i&lt;n*6and-~f(n,i+1)


The formula n=i*(i+1)*(i+2)/6 for tetrahedral numbers can be more nicely written in i+1 as n*6=(i+1)**3-(i+1). So, we find the lowest i for which i**3-i&lt;n*6. Each time we increment i starting from 1, the recursive calls adds 1 to the output. Starting from i=1 rather than i=0 compensates for the shift.