Postgresql – What would be an efficient way of preventing sequential scan if data is partitioned by list, when it comes to updating multiple partitions

I hate to state the obvious but I'd say test both scenarios and run your production queries against all 3 scenarios (scenario 1 being the current non partitioned one). The reason I state this is because I don't know what your code is querying. Do they actually have a benefit of being sorted in the base table by the identity column as opposed to the other column? For example, are your queries actually looking for the row ID often? If you're not sure, you might be surprised by getting some performance benefits. The general idea of using ID as the clustered key was for range scans, but in our case, we scan by customerID and Date, so it worked out perfectly for us, and perhaps you. Check out this article by Kim Tripp: http://www.sqlskills.com/blogs/kimberly/post/The-Clustered-Index-Debate-Continues.aspx

"What is often cited as the “reason” for IDENTITY PRIMARY KEY clustered index definitions is its monotonic nature, thus minimizing page splits. However, I argue that this is the only “reason” for defining the clustered index as such, and is the poorest reason in the list. Page Splits are managed by proper FILLFACTOR not increasing INSERTS. Range Scans are the most important “reason” when evaluating clustered index key definitions and IDENTITies do not solve this problem.Moreover, although clustering the IDENTITY surrogate key will minimize page splits and logical fragmentation due to its monotonic nature, it will not reduce EXTENT FRAGMENTATION, which can cause just as problematic query performance as page splitting. In short, the argument runs shallow "

Typically you want your clustered index to be as narrow as possible, unique, and non nullable as it get's carried into all other indexes. I have a table with roughly 10 billion rows and we partition off the datetime column, which has worked out great.

Only rows in 7 partitions qualify the predicate(s) so there is no need to ever lookup rows in the other 3 partitions.

The query plan may choose to scan on B and probe (lookup) on A.

A Key Lookup is always a seek into the clustered index. There is no partition elimination in your original example by the way - the query processor is just reporting that it touched 7 distinct partitions (3..9) while executing the query. None of the lookups accessed other partitions.