NO
The way I read the specification:
Rprog = 1000 * Vprog/ Ibat
for 500 ma (aka 0.5A)
Rprog = 1000 * 1/.5 = 2k Ohms
However if Vprog is on the high side of the specification (aka 1.07) the battery charge current is:
Ibat = Vprog/Rprog * 1000 = 1.07/ 2k *1000 = 535 ma