Double partition of list

Partition can have lots of different arguments so I am curious whether the following results can be achieved using single Partition instead of two.

Example 1:

Partition[#,2,1]&/@Partition[Range[12],3]

(* {{{1,2},{2,3}},{{4,5},{5,6}},{{7,8},{8,9}},{{10,11},{11,12}}} *)


Example 2:

Partition[#,2,1]&/@Partition[Range[20],4]

(* {{{1,2},{2,3},{3,4}},{{5,6},{6,7},{7,8}},{{9,10},{10,11},{11,12}},{{13,14},{14,15},{15,16}},{{17,18},{18,19},{19,20}}} *)


You need to have nested lists in order to have partitions with different offsets at different levels.

1st example

Starting with:

pts0 = Partition[#, 2, 1] & /@ Partition[Range[12], 3]

pts0sol =
First@Partition[{Range[1, 11], Range[2, 12]}, {2, 2}, {2, 3}]

pts0 == pts0sol


True

2nd Example

pts1 = Partition[#, 2, 1] & /@ Partition[Range[20], 4]

pts2sol =
First@Partition[{Range[1, 20], Range[2, 21], Range[3, 22]}, {3, 2}, {2, 4}]

pts1 == pts2sol


True

In my opinion, this would become less readable with increased nesting complexity.