Sezuh,
You don't need to have the complete schedule of 49 rounds to verify that it will work. Consider the 50th player represented by infinity who will always occur in the first fivesome, the other four players in that fivesome cycle through all the numbers 0..48, so it is easy to see that the 50th player occurs exactly 4 times each with the other 49 players.
The 49 players who are represented by the numbers mod 49, have 49*48/2=1176 possible pairs among themselves. These pairs can be divided into 24 subsets of 49 pairs, where each subset contains all pairs of players who have the same difference modulo 49. Consider the pair (5
which occurs in the second fivesome and the pairs that will be developed from it over the next 48 rounds by adding one each time. These will be (5,8),(6 9),(7 10),..,(45 48),(46 0),(47 1),(48 2),(0 3),(1 4),..,(4 7). These are all of the pairs that differ by 3 mod 49. The same subset of 49 pairs, but in a different order, will also be generated by the pair (13 16) in the 4th fivesome, the pair (39 42) in the 8th fivesome and the pair (17 20) in the 9th fivesome. So I have verified that the complete schedule will contain these 49 pairs at least 4 times each. To show the complete schedule is fully balanced, you just need to examine the differences of all the pairs taken from the 10 starter blocks and verify that all 24 differences mod 49 are covered exactly 4 times each.
Hope that helps.
Ian.