With visible light, it is definitely much better to use DC drive, but when building sensor "systems", pulsing the LED gives you a definite S/N benefit.
https://www.sbprojects.com/knowledge/ir/ir.htm
In any case, it looks like most IR protocols end up having 25% duty cycles or less - the 38KHz modulation gives you 50%, and the bi-phase gives you another 50% multiplier. Pulse durations are 1/38KHz = 2.6E-2 msec. This means that the part is specified for ~400mA or so. Obviously recheck this if you plan on doing some other protocol.
Incidentally, about putting the diodes in parallel, there seems to be a ~.4V spread on the forward voltages. At a current of 400mA, the forward voltage is ~2V, and if you look at the V/I curve, there looks like a knee in the plot. Looking at the slope of the curve at that point, a .5V difference can cause a .4A change. This .5V/.4A=1.3 Ohm "dynamic resistance" will balance out the currents a little bit - but putting in little more resistance is a good idea. A +-.2V difference roughly translates into a (.2V / "1.3Ohm") = +-150mA difference in currents.
And finally, power dissipation: 400mA * 2.4V (worst case) * 25% (duty cycle) = ~250mW which is actually above the maximum power dissipation of the device. In normal use, there are delays between pulses, so this is probably okay, but at these settings, the thermal resistance value predicts a pretty insane (.24W*350K/W) = 84Deg. C temperature rise over ambient - i.e it's going to smoke if it continously transmits. To keep the junction temperature under 100C, we should limit the temperature rise to ~60C or so - so we should somehow scale the power back by (84-60/84) = ~30%, so one quick fix is to take the 400mA number and push it down to ~300mA. Obviously this is conservative - LED forward voltage also goes down a couple tenths and you can heatsink the LED a bit as well - but it's guaranteed to work.
James