$1^3 , 2^3 , ... , n^3$

Discussed here.

generalisation: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=348347&start=0 see #2 and #4

For $n=0$ (mod 3) take $n$ times number $n$: $$ Prod=n^n, \ \ Sum=n^2. $$ For $n=1$ (mod 3) take $(n-4)$ times number $(n+1)$, number $\frac{5n+4}{3}$ three times, and $1$ one time. Those numbers are $$ n+1,\dots,n+1, \,\frac{5n+4}{3},\frac{5n+4}{3},\frac{5n+4}{3}, \,1. $$The product and the sum are $$ Prod=\left(\frac{5n+4}{3} \right)^3 (n+1)^{n-4}, \ \ Sum=(n+1)^2. $$ For $n=2$ (mod 3) we take $n+2$ times number $n-5$, eight times number $1$, and $\frac{5n+2}{3}$ three times. The numbers are $$ n-5,\dots,n-5, \, 1,1,1,1,1,1,1,1,1, \, \frac{5n+2}{3},\frac{5n+2}{3},\frac{5n+2}{3}. $$The product and the sum are $$ Prod=(n-5)^{n-2} \left(\frac{5n+2}{3} \right)^3, \ \ Sum=(n+1)^2. $$