The Saros cycle
To elaborate on L.Dutch's answer, a saros is the unit of time in which a certain Sun-Earth-Moon geometry will (almost) perfectly repeat itself. For our Moon and Earth, this number is equal to 6585.3211 days (or 18 years and 11.321 days). That means if you had an eclipse (either lunar or solar) on a particular day, then you would also have an eclipse one saros (6585.3211 days) later. Each successive eclipse after a saros is called the saros series. Note that the saros only tells you when the next eclipse occurs given you have already had one. To calculate eclipses from scratch, you would need an orbital mechanics propagator, which is beyond the scope of this (already long) answer.
Since there are more than one geometry during a year that can cause an eclipse, there is more than one saros series. You can see all the saros series here. If you look at that list of series, you will see that each series has a "first eclipse" and "last eclipse". That's because the saros cycle isn't perfect. Eventually they stop and end. They tend to be active for at least ~500 years, spanning ~70 eclipses.
Note that a saros is not a whole number of days, therefore the Earth spins 6585 and a third times. This extra 1/3 of a spin means the next eclipse in the saros series won't occur at the same location on Earth; instead it occurs about 1/3 a revolution (120 degrees) in longitude to the west. For a particular eclipse then, it requires three saros to repeat at the same location on Earth.
What this means for a fictitious planet: calculate a saros
The saros cycle is the number of days that give (near) whole periods to the Moons synodic, anomalistic, and draconic periods. These months are:
- Anomalistic Month $T_a = 27.55455$ days. The time the moon takes to come back to a particular point in it's orbit (perigee usually). This is the newtonian-esque period that most people mean when they talk of lunar periods, and can be calculated from Kepler's Law: $$(T_a / 27.55455 )^2 = (a / 384784 )^3$$ where $T_a$ is in days and $a$ is in kilometers.
- Synodic Month $T_s = 29.53059$ days. The period the moon takes to come to the same position relative to the Sun. This is different since the Moon orbits the Sun in a prograde orbit. You can estimate it from the Anomalistic Month as: $$T_s = T_a (1 + \frac{T_a}{Y})$$ where $Y = 365.25$ days for Earth. Substitute your planet's year.
- Draconic Month $T_d = 27.21222$ days. The time it takes the Moon to return to either an ascending or descending node. Since the eclipses only occur when the Moon is at one of these nodes, it must be included in a saros. It differs from the anomalistic month because the nodes precess (move ever so slightly). Calculating this will take us way off course, so lets just assume it has the same proportion to the anomalistic Month as Earth's does: $$ T_d = \frac{27.21222}{27.55455} T_a $$
Your goal then is to calculate $T_a$, $T_s$, and $T_d$ for your hypothetical system. Next, find the number of days that produce (near) whole numbers of periods. There's no equation to do this; you will need a spreadsheet or a computer program. Here's the algorithm:
set n = 1
set limit = 0.25
Do the following until error <= limit:
saros_a = n * T_d
saros_s = round( saros_a / T_s) * T_s
saros_d = round( saros_a / T_d) * T_d
error = largest saros - smallest saros
increment n by 1
return saros_a
This will calculate Earth's saros number, and so should be good enough for a fictitious planet. Once you have your saros number, sprinkle a couple of starting eclipses in around your planet's equinoxes and get to calculating. Good luck!