The discharge slope, or output low time, only depends on R3 when R3 is too low. It's also worth noting that R1 and R3 in your schematic form an equivalent resistance. Is there a reason they are separate? Is R1 fixed and R3 a pot in practice? If not, you can just use one resistor.
From the wikipedia article, "Particularly with bipolar 555s, low values of \$R_1\$ must be avoided so that the output stays saturated near zero volts during discharge, as assumed by the above equation. Otherwise the output low time will be greater than calculated above."
In normal operation, the voltage drop inside the 555 timer's discharge pin is fixed while the capacitor is discharging. The bulk of the voltage drop is across the R1||R3 resistance. Thus, the discharge pin is at nearly 0V and the capacitor discharges at the normal rate. However, when the R1||R3 resistance is too low, there is less voltage drop across these resistors and more internal to the timer. Then, the discharge pin is no longer near 0V and the capacitor takes longer to discharge. When R3 = 0, the same thing is happening, just to an extreme. There is theoretically no voltage drop across R3||R1. Instead all the voltage drop from Vcc to GND is internal to the 555 timer. Thus, the capacitor is simply held at about Vcc.
As the other suggested, it's good to look at the internals of the 555 timer to understand what is really going on. I've had to do the same in the past. These java applets are a good way to play around with circuit and see what is really going on.
http://falstad.com/circuit/e-555int.html - 555 Internals
http://falstad.com/circuit/e-555square.html - Astable Multivibrator
