Led lighting in "2 steps" (555 blinker circuit)

I've seen damaged LEDs do this.

Also note that Ra=Rb does not give you 50% duty cycle. It gives you about 33% (or 66% - depending how you have the output wired). Can you link to your circuit so I can be sure?

Also 1M is getting a little large for resistors used with a 555. Try 100k and a 10uF capacitor.

 

Ah, OK. The 555 is pretty noisy when it switches. The Glitch on the supply can cause other logic to misbehave. In this case it's apparently causing the 555's comparators to mis-trigger. The capacitor on pin 5 maintains the input voltage on the comparators even if the power supply voltage goes glitchy.

The specs say that "maximum value is R = RA + RB ≉ 3.4MΩ, and for VCC = 15 V, the maximum value is 10MΩ." see here.

The reason is the threshold current. As the resistor values increase, the current flowing through this pin can exceed the current flowing through the resistor. It's best to keep well away from this as it will affect the timing before it stops it.

The specs are assuming the worst case for this, so it's almost always safe. You just don't want to be bitten by it on the rare occasions it doesn't. If you can keep away from it you'll be safer.

In your case Ra + Rb is not an issue because of your diode. In your case it's only Ra that's important, and 1M << 3.4M.

I tend to keep the Ra + Rb under 1M if possible.

 

It likely won't make any difference, but remember that if you divide the resistors by 10 you need to multiply the capacitor by 10 (so 10uF)