Stop. Just stop.
You need to learn about driving LEDs. Apparently what you want to do is
simulate this circuit – Schematic created using CircuitLab
where the number of LEDs will depend on the voltage of V1 and the color of the LEDs in a string. Am I correct?
If so, you are in big trouble. What will happen is that the LEDs will light up to some brightness. Very likely, the current being drawn will be too great, and you will kill one or more LEDs. You must keep in mind that the data sheets mention a range of values for things like forward voltage, and you cannot count on getting a bunch of "typical" units - not unless you're willing to fix things after they go badly wrong. But let's say that you've gotten lucky, and the LEDs are drawing just the right amount of current. What then?
Well, the LEDs will start to heat up. When they do, two things will happen. First, they will get dimmer. Second, they will begin to draw more current at the fixed voltage. This will cause them to get even hotter, which will cause them to draw even more current, which will cause them to get even hotter... and eventually one LED will die. If you are lucky, it will fail open circuit, which will save the others even though the string will be dead. If you are not lucky, it will fail shorted, and the situation will get much worse very quickly, and a second will fail. If that one fails shorted, a third will die very quickly indeed, and the process will continue until all are dead or the power supply decides it's had enough and shuts down. The increase in LED temperature is called "thermal runaway", and the system failure is called "firecracker mode".
So, how do you avoid this? Simple, you don't connect the LEDs directly to a voltage source. The simplest way is to add a series limiting resistor to the LED string, like so
simulate this circuit
How big a resistor, you ask? At a minimum, I'd recommend a resistor which drops 1/4 of the power supply voltage, although I'm conservative in these things. So, if you were planning to drive a string of LEDs at 2 amps from a 36 volt supply, you could go with 4.5 ohms, which will produce a 9 volt drop. The value is not super-critical, so you could get away with a 5 ohm resistor, which is a common resistance value. Of course, then you'd only have 27 volts available for your LEDs, but that's tough. Also, 2 amps at 9 volts is 18 watts, so you would need at least a 20-watt resistor, and a bigger unit would be better (Did I mention I'm conservative on these things? Did I also mention that I've flown rocket payloads, including a satellite, and my boxes have never failed? There is probably a connection there.).
This is obviously inefficient. The alternative is to use a switch-mode current source, which will turn the LED on and off very quickly, and get a proper average value. It will do so with a relatively small voltage drop, which means you can use more of your voltage for the LEDs. It's pretty apparent that you are not at the stage of being able to design this sort of circuit, so I really don't recommend that you try it just yet. You can buy commercial LED drivers which do operate this way, and if you're willing to take your chances you can get them very cheaply on eBay. Keep in mind that the cheap ones come directly from unknown companies in China, and the old saying, "caveat emptor" ("buyer beware") applies. It is your choice, though.
But let's say you've decided to go ahead with the DIY approach. Now you need to size your heat sink. You have stated that you want to mount a total of 262 watts worth of LED on each heat sink. If we make the generous assumption that the luminous efficiency of the LEDs is 20%, then we can expect to dissipate 80% of the input power as heat, or 210 watts. Additionally, we can expect to lose 1/3 of the LED power to the series limiting resistor, or another 83 watts. Total heat that the heat sink must handle is 290 watts. If you want the LED temperatures to be no more than 30 degrees C above ambient (which will put them up at about 130 degrees Fahrenheit), you will need a heat sink with a thermal resistance of 0.1 degrees/watt, since 0.1 times 300 is 30. Actually, you will almost certainly need a separate heat sink for each LED, since you'll want to space the LEDs out more than a single heat sink will allow. You can get such heat sinks on eBay, and they will run you about 1 dollar each, or maybe a bit more. You will also need to provide heat sink for each limiting resistor. If you do try to go with a single heat sink for each 262 watts of LED power, you'll find that it will run you somewhere near 100 bucks or more, and will be big and heavy, AND you will absolutely need a cooling fan for each heat sink.