In the last question ( [Numbers of the form $\frac{xyz}{x+y+z}$](https://math.stackexchange.com/questions/2378476/numbers-of-the-form-fracxyzxyz) ), I asked if every natural number $n$ can be written as 
$n=\frac{xyz}{x+y+z}$
with $x,y,z\in \mathbb{N}$.
A few answers were given with some formulas which show that this is always possible.
I collected those formulas and checked with some small SAGE program, which is given below, if in the range $1 \le x,y,z \le 10$ every solution found can be explained by any of the formulas. Unfortunately this is not the case:
 Here are some counterexamples not covered by the formulae $(n,(x,y,z))$:

```
 (14, (8, 7, 5)),
 (20, (5, 10, 10)),

```

Here are the formulae so far:

1. If $n=a\cdot b$, then $x=2b, y=2b+a,z=a$
2. $x = n+1, y = (n+1)^2-1, z=1$
3. $x = 1, y = n+2, z = n(n+3)/2$
4. for $n\equiv 1 \mod(2)$: $x=3,y=\frac{n+3}{2},z=n$.
5. if $n+1=ab$, then $x=n,y=a+1,z=b+1$ (Ronald Blaak)
6. if $n=3a^2$ then $x=y=z=3a$ (PJF49)
7. if $n=a^2$ then $x=a,y=a+1,z=a(2a+1)$ explains 4, (2, 3, 10)
8. if $n\equiv 2 \mod(3)$ then $x=2,y=2n, z = \frac{2(n+1)}{3}$ explains 5, (2, 4, 10)
9. if $3n=ab$ and $a\equiv b \mod(2)$ then $x=a,y=\frac{a+b}{2},z=b$ explains: (8, (4, 5, 6)), (16, (8, 7, 6)), (20, (6, 8, 10)), (21, (7, 8, 9)), (15, (9, 7, 5))

Here is the SAGE code if you want to run it on your own:

```
from sage.all import *
from collections import Counter

def isSolution(t,sol):
  a,b,c = t
  X,Y,Z = sol
  return (a,b,c) in [(X,Y,Z),(Y,Z,X),(Z,X,Y),(Y,X,Z),(Z,Y,X),(X,Z,Y)]

N = 10
foundSolution = []
s = []
for x in range(1,N+1):
  for y in range(1,N+1):
    for z in range(1,N+1):
      if Mod(x*y*z,x+y+z)==0:
        n=(x*y*z)/(x+y+z)
        solutions = []
        for d in divisors(n):
          a = d
          b = n/a
          xx = 2*b
          yy = 2*b+a
          zz = a
          solutions.append(((xx,yy,zz),"divisor"))
        for a in divisors(n+1):
          xx = n
          yy = a+1
          zz = (n+1)/a+1
          solutions.append( ((xx,yy,zz),"divisor of n+1"))
        solutions.append(((n,n+2,2),"n+2"))
        solutions.append(((n+1,(n+1)^2-1,1),"n+1"))
        solutions.append(((1,n+2,n*(n+3)/2),"1,n+2"))
        if n%2==1:
          solutions.append(((3,n,(n+3)/2) ,"n%2==1"))
        solutionFound = False
        for sol,t in solutions:
          if  isSolution((x,y,z),sol):
            solutionFound = True
            foundSolution.append((sol,t))
        if not solutionFound:
          print n,(x,y,z)
          s.append((n,(x,y,z)))

print Counter([s[1] for s in foundSolution])

```

**If you happen to find another formula, which is not on the list *and explains one of the given numbers*, then please write your solution, so that I can add the formula to the list.**