Here is the molecule. Clearly the 4th carbon is chiral. I have drawn the diagraph for it, but the exploratory pathways of the left and right chains are the same. So, the problem is that the two chains are actually different, but same in the diagraph. How to assign stereodiscriptors in such cases?
Not the proper, somewhat complicated, IUPAC ‘digraphs’… but might help.
Bold/colored bonds are the left and right branches' resulting paths taken when advancing to next sphere (because the decision could not be yet done) during determining a higher priority branch by finding a first difference of the compared atom set rankings. Sphere numbers a Roman. Grey digits are semi-conveniently arbitrarily chosen numbering just for reference, not nomenclature numbering nor priorities(!).
Following list shows chosen highest priority atoms in comparison in each nth sphere and left/right branch (bold; also comparison with other sibling atoms in the same branch, if they are not just hydrogen), chosen according to the next bonded atoms from the (n+1)th sphere (in 1st braces; atom number in 2nd brace).
sphere № : [left] vs [right]
I: [ C(CCH) (1) ] = [ C(CCH) (4) ]
II: [ C(CCH) (2) > C(CHH) (3) ] = [ C(CCH) (5) > C(CHH) (6) ]
III: [ C(HHH) (7) = C(HHH) (8) ] < [ C(OHH) (9) > C(HHH) (10) ]
Please note that in the II. sphere, left, the carbon (3) leading to terminal -OH group could not be chosen, because it's C(CHH), compared to higher priority, chosen C(CCH) (2).
Also note that in sphere III, left, any of the two methyls' carbons could be chosen (equivalent).
In this III. sphere, the decision finally could be made. The right branch (R1) has higher priority – finally only hydrogens bonded at left: C(HHH) (7), vs oxygen and hydrogens bonded at right: C(OHH) (9).
(Next priority is left branch R2, then methyl, then H.)
So the compound configuration/name is