[Math] Summing of factorials to produce perfect cubes

I was playing around with factorials the other day, and I realized that $4!+5!$ is a perfect square. Perplexed by this result, I started looking for other pairs of factorials that produce a perfect square when added together (unbeknownst to me, I had stumbled across a well-known open problem in number theory). I could only find three: $1!+4!$, $1!+5!$, and $1!+7!$. Since $4!+5!$ seemed like an outlier, I decided to focus on the three which satisfy the formula $n!+1=m^2$. I turned to the main Math.SE chatroom, asking if there were any more, which is when I was informed that this is called Brocard's problem. Since it's a well-known open problem, I decided to put my own spin on it. It has probably been done before, but I can't find anything online.

What about the equation $n!+1=m^3$? As far as I know, there is no pair of integers that satisfies this equation. What if the conditions are relaxed? This brings me to the following question I came up with yesterday:

How many perfect cubes can be expressed as the sum of two or more unique integer factorials without subtractions, divisions, or multiplications of factorials?

Notice I said unique, meaning any factorial cannot appear more than once in a sum. With that in mind, I've only found three:$$ \begin{align}&2!+3!=8=2^3\\&1!+2!+4!=27=3^3\;\;\;\Bbb{and}\\&1!+2!+3!+6!=729=9^3\,.\end{align}$$Are there more solutions, or are these the only three?

Using this link, I've concluded any additional solution must have the factorials add up to be greater than $341^3$. Since I hardly have any experience with programming, I can't go much further on this problem… Any and all help would be greatly appreciated.

Note: My question is NOT a duplicate of this one. I'm not restricting my problem to the partial sums $S_n$ of the series $\displaystyle \sum_{k=1}^nk!$.

jagy@phobeusjunior:~$ ./greedy_factorial Sat Dec 13 12:25:29 PST 2014 0 1 1 8 3 2 9 3 2 1 25 4 1 27 4 2 1 32 4 3 2 121 5 1 128 5 3 2 144 5 4 729 6 3 2 1 841 6 5 1 5041 7 1 5184 7 5 4 45369 8 7 3 2 1 46225 8 7 6 5 4 1 363609 9 6 3 2 1 403225 9 8 4 1 3674889 10 8 7 6 3 2 1 1401602635449 15 14 13 12 11 10 9 8 7 3 2 1 Sat Dec 13 12:26:23 PST 2014

jagy@phobeusjunior:~$ ./greedy_factorial 0 square root 0 = 1 square root 1 = 1 8 = 2^3 9 square root 3 = 3 25 square root 5 = 5 27 = 3^3 32 = 2^5 121 square root 11 = 11 128 = 2^7 144 square root 12 = 2^2 3 729 square root 27 = 3^3 841 square root 29 = 29 5041 square root 71 = 71 5184 square root 72 = 2^3 3^2 45369 square root 213 = 3 71 46225 square root 215 = 5 43 363609 square root 603 = 3^2 67 403225 square root 635 = 5 127 3674889 square root 1917 = 3^3 71 1401602635449 square root 1183893 = 3 394631

ALLOWING 0! = 1 ***+++*** jagy@phobeusjunior:~$ ./greedy_factorial Sat Dec 13 11:46:03 PST 2014 0 eleventh root 0 1 eleventh root 1 4 square root 2 8 cube root 2 9 square root 3 25 square root 5 27 cube root 3 32 fifth root 2 121 square root 11 128 seventh root 2 144 square root 12 729 cube root 9 841 square root 29 5041 square root 71 5184 square root 72 45369 square root 213 46225 square root 215 363609 square root 603 403225 square root 635 3674889 square root 1917 1401602635449 square root 1183893 Sat Dec 13 11:46:36 PST 2014 jagy@phobeusjunior:~$

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EDIT: Wrote a command to do sum of distinct factorials by greedy algorithm; here are the perfect powers I have, not allowing $0!$.

Did not get any more cubes yet.

Allowing $0!$ adds in the square $4,$ I'm not sure i see any other change. Not sure why the program is so very slow. As I said, if all you want is cubes, you can program in a greedy algorithm which, for each cube, will rapidly attempt a decomposition as distinct factorials (with repeat 1 for $0!$ if you wish ) and report success.

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