I think you need 1.57 times less for anhydrous, so 1.27 g.
This is correct.
In the interest of education, I am a retired chemistry professor. ;-)
Here is how you figure this out:
First a couple of definitions... "anhydrous" means "without water" and "pentahydrate" means "with five waters".
The formula weight of sodium thiosulfate (Na
2S
2O
2) is 158.11 g/mol.
The formula weight of water (H
2O) is 18.02 g / mol. Therefore, five waters are 90.1.
Thus the formula weight of sodium thiosulfate pentahydrate is 158.11 + 90.1 = 248.21 g/mol
Therefore the "conversion factor" between the two is 158.11 / 248.21 = 0.637. (The value cited by iakustov is the reciprocal of this value.)
In other words for every gram of the pentahydrate your recipe calls for use 0.64 grams of the anhydrous form.
Regards,
--- Frank