For the series resistor you must use the following value:
R= (Vbatt-Vleds) / Ileds.
Vbatt is the battery voltage
Vleds is the total forward voltage of the leds
Iled is the wanted current through the leds.
So in your case the resistor will be for a 10 mA current:
(4.5 - 4.2)volts / 0.01 Amp = 30 Ohms
When you remove one led the current will be:
(4.5 - 2.1) Volts / 30 ohms = 0.08 Amp , wich is 80 mA , wich will likely blow the led.
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