Let's see. The device size is 1,465,138,583½ kB = 1,500,301,909,504 B. The filesystem consists of 366,284,288 blocks of 4096 B each, which is 1,500,300,443,648 B. I don't know what the remaining 1,465,856 B (1.4 MB) are used for (additional copies of the superblock? I know there are a few kB of space at the beginning for the bootloader.).
The filesystem contains 91,578,368 inodes of 256 bytes each, which takes up 23,444,062,208 B (about 22 GB, hint, hint). Then there is 1,442,146,364 kB = 1,476,757,876,736 B for file contents. This accounts for 23,444,062,208 B + 1,476,757,876,736 B = 1,500,201,938,944 B. The remaining size is 98,504,704 B = 24,029 blocks which is in the right range to be the journal size.
As you can see, everything is accounted for. (Ok, almost everything, but we're talking megabytes, not gigabytes.)
Best Answer
mdadm supports dealloc.
commit=secis the time, the filesystem syncs its data and metadata. Setting this to 0 has the same effect as using the default value 5.So I don't get the link between mdadm and commit=0 in your question?