Turning on the LED with no resistor does not mean there is no current limiting. It just means that current is limited by:
- internal resistance and inductance of the LED
- the same of your driving circuitry
- your power supply's ability to supply current
If your power supply is a couple of AA batteries, then it probably can't supply much current. If your power supply is a 12V automotive lead acid battery, your LED will probably be reduced to cloud of metal and plastic vapor before you even realize what's happened.
Determining the current and duty cycle can be done by a rough rule of thumb: brightness is proportional to average power. So, if the LED has the brightness you like at a constant \$10mA\$, then it will have about the same brightness at \$20mA\$ pulses and a 50% duty cycle, or \$40mA\$ pulses and a 25% duty cycle.
There are two problems here. Firstly, when you start pulsing, even if you hold the average power the same, the peak power increases. This means the peak temperature increases also, and if the LED isn't given sufficient time for the heat to dissipate through the device between pulses, it will be damaged. This is what limits the peak current.
The maximum peak current will be given in the LED's datasheet. Usually, what limits the current is the LED's ability to dissipate heat. See this Cree application note for more on that: Pulsed Over-Current Driving of Cree XLamp LEDs: Information and Cautions. Of course, these are high-power LEDs. Small indicator LEDs will be more fragile.
The other problem is that as current increases, \$I^2R\$ losses increase also, and the LED becomes overall less efficient. See Does pulsing an LED at higher current yield greater apparent brightness?
PFC on switch mode power supplies would not be very effective if the dc output dropped to zero every ten milli seconds. This of course is avoided by using capacitors to store energy so that PFC works as expected.
I'm not sure about the chip you are using but it would seem a trivial addition of a couple of components would all that is needed to avoid LED flicker.
Best Answer
Ignoring the effects of contamination in the medium between the emitter and the receptor, the energy received when the receptor is up-close to the emitter will be the same as the energy received when the receptor is at a greater distance. This of course assumes that the receptor has a big enough surface area to receive the "spread-out" light.
What goes down is the received energy per square metre (as the distance is increased) but, nevertheless, the same total energy will hit the receptor because as the energy per metre thins-out, it illuminates a bigger surface area hence, energy is the same.