A curiosity on a first three natural numbers Let's review a triple of numbers, $1, 2, 3$, it is a curiosity that 
$$1+2+3 = 1\times2\times3 = 6$$
Are there another triples (or not necessary triples) such that their multiple equal to their sum?
And generalised pattern of such identities would be interesting and appreciated.
PS: Conjecture: Reviewing $t$ fold case of such numbers, they are seem to be the integer solutions of the equation
$$n(n+1)(n+2)\cdots(n+t) = \binom{t+1}{1}n + \binom{t+1}{2}$$ 
PSS: Integer solution (for consequent integers)
$$\prod_{k=0}^{2s} (n+k) = \sum_{k=0}^{2s} (n+k)$$ 
for $n=-s$. But these sums and products are 0.
PS3: Still we can easily find such combinations using the following pattern:
$$\prod_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1 \times \prod_{k=0}^t a_k = \left(\sum_{k=0}^t a_k\right)+\sum_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1$$
 A: Working with integers
$$n(n+1)(n+2)=3n+3=3(n+1)$$
With $n=-1$, we have $$-1,0,1$$ as a solution
Otherwise 
$$n(n+2)=3$$
$$n^2+2n-3=0$$
$$(n+3)(n-1)=0$$ 
$$n=3,n=1$$
Thus we have $$-3,-2,-1$$ or $$1,2,3$$ as solutions. 
A: "And generalised pattern of such identities would be interesting and appreciated"
Well, while there is nothing particular about the triplet $59,60$ and $61$, we do have $tan59+tan60+tan61=(tan59)(tan60)(tan61)$, the triplet being in degrees. This particular case comes from the identity $tanA+tanB+tanC=(tanA)(tanB)(tanC)$ where $A+B+C=180$.
A: If we know, that $A=1,B=2,C=3$ is a solution we can look for another solution with larger numbers by
$$(A+a)+(B+b)+(C+c) = (A+a)(B+b)(C+c)  \\
-----------------------------\\
(1+a)+(2+b)+(3+c) = (1+a)(2+b)(3+c)\\
6+a+b+c = 6+ 2c+3b+6a+bc+3ab+2ac+abc\\
a+b+c = 2c+3b+6a+bc+3ab+2ac+abc\\
0 = c+2b+5a+bc+3ab+2ac+abc\\
$$
If no number $a,b,c$ is negative, all must be zero.
A: Hint: $$a+b+c=abc$$ and $a<b<c$ , so we have $$3c>abc\implies ab<3$$
So since $ab >1$, we have $ab=2$, so $a=1$ and $b=2$ and ...
A: 
Are there another triples (or not necessary triples) such that their multiple equal to their sum?

Easy. Just take some random numbers, say $3$ and $4$. We have $3 \cdot 4 = 12$, but $3+4=7$. So, just pad it with $12-7=5$ more $1$'s, and we have:
$1+1+1+1+1+3+4=1\cdot1\cdot1\cdot1\cdot1\cdot3\cdot4=12$ 
A: Considering $(n,n+1,n+2)$ solutions with $n\in N$, I have: $3n+3=n(n+1)(n+2) \Leftrightarrow 3(n+1)=n(n+1)(n+2) \Leftrightarrow 3=n(n+2)$. So, I obtain: $$n^2+2n-3=0$$The equation becomes: $(n-1)(n+3)=0$ so $n=1$ or $n=-3$. The second solution is impossible because $n \in N$, the first is the only correct.
