Hello vSohill,
Not sure exactly what you mean but guessing you are asking about FTT=X with FTM=RAID1 vs RAID5/6, here is breakdown of how this is composed:
FTT=0, FTM=RAID0 = 2(0)+1 = 1 component (1 data-replica)
FTT=1, FTM=RAID1 = 2(1)+1 = 3 components (2 data-replicas + 1 witness component)
FTT=2, FTM=RAID1 = 2(2)+1 = 5 components (3 data-replicas + 2 witness components)
FTT=3, FTM=RAID1 = 2(3)+1 = 7 components (4 data-replicas + 3 witness components)
FTT=0, FTM=RAID5 = doesn't exist as RAID5/6 doesn't work like this on vSAN nor on any storage platform
FTT=1, FTM=RAID5 = 4 components (all components contain a combination of data and a single set of distributed parity data)
FTT=2, FTM=RAID6 = 6 components (all components contain a combination of data and two sets of distributed parity data)
So no, there is no formula for RAID5/6 minimums other than minimum 4/6 node/Fault Domains needed for FTT=1, FTM=RAID5 and FTT=2, FTM=RAID6 respectively.
Bob