I'd plan to have one single resistor feeding the parallel branches of LEDs.
In theory, with V=IR, that would mean I have 4=0.9*R, so I'd need a resistor of 4.5 ohms.
That means that via W=AV, I'd need to handle 0.9A * 2V = about 2 Watts. So I'd probably bump that up for some buffer and need a 4.5 ohm, ~5 Watt resistor.
So if I plan for 10mA per LED, then I'd only use 300mA of steady current. That means 4=0.3*R, so R=13.3 Ohms, and wattage could drop to about 2 with padding (e.g. 0.3A*2V = 0.6W).
Since AAs have mAh of around 2.5
That would mean I'd have to roughly take the current down to 1mA per LED, which I can't imagine would be enough.
I'm a doofus - everywhere I used 4, I meant 3. Because 2 AA batteries in series should be 3 Volts, I think - 1.5 + 1.5
To Deva - WRT old phone charger - I'm trying to avoid plugging into AC power. I thought about putting as many LEDs in series as possible so they could light up sufficiently, but with 2 AA in series and a ~2V forward voltage drop, at most I could do 2 LEDs in series
Note that in the data sheet it says Vf is typically 1.85V with a maximum of 2.5V. This is why you can't put them in parallel with one resistor feeding all of them, the ones will the lowest Vf will take all the current
If I use this 3MM LED from Digikey (Datasheet: https://www.lumex.com/spec/SSL-LX3054SRD.pdf), then the steady current would be 30mA * 30 LEDs = 900 mA of steady current.
Second question -- clearly 900mAs can't be sourced realistically from 2 AA batteries.
But since AAs have mAH of around 2.5, then I could run for about 8 hours before running dry. That's not enough. I figure the batteries need to last for 3-4 weeks realistically, of about 5 hours per day. So if I could get to ~100 hours I'd be good.