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:
$$\prod_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1 \times \prod_{k=0}^t a_k = \left(\sum_{k=0}^t a_k\right)+\sum_{k=1}^{a_0\cdots a_t - (a_0+\cdots+a_t)} 1$$

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.
