I Led is the current through (used by) the Led.
Since two LED should pass/use pretty much twice the current, you would alway divide through twice as much for 2 LEDs so the result should always be half compared to one.
So, correct me if I'm wrong, but as far as I understand you would simply use a resistor with half as much ohms for two LEDs in parallel.
Here are a few example calculation, guesstimating the supplied V and current through LED:
- 2/3V (2.5V) LEDs with 470 Ohm resistors as mentions on GONs website
- Guesstimating a current of 20ma/Led (which seems like a fairly realistic/common current for LEDs)
-> Results in a supplied Voltage of 12v (which seems unrealistic with usb being only 5v)
- 2/3V (2.5V) LEDs with 470 Ohm (500 Ohm) resistors as mentions on GONs website
- Guesstimating a supplied Voltage of 5v (since usb is 5v)
-> Results in a current of 5ma/LED