The Josephus Problem - Numberphile

In Round II, #1 starts by bumping off #3. Thus, the numbers that survive correspond to (1+4n) and therefore, the last one surviving the round is (1+4*24) ie 97 who proceeds to bump off #99 and hands the dagger to #1.

So far so good?

In Round III, #1 bumps off #5 and hands the dagger to #9. Survivors follow the pattern (1+8n) with #97 being the last one in the round (1+8*12). The person next to him is #1 and it is his turn to get bumped off. Next round starts with #9

In Round IV, survivors are {1+8*(2n+1)} - starting with #9 who bumps off #17 (1+8*2) and hands the dagger to #25 (1+8*3). The last survivor of this round is (1+8*11) #89 who kills #97 and hands the dagger to #9.

In Round V, survivors are #9 {(1+8*1)}, #25 {(1+8*3)}, #41, #57 & #73. Alternate knocks and #9 kills #25, dagger to #41 who kills #57 who gives dagger to #73 who kills #89 and hands the dagger over to #9.

In Round VI, survivors are #9, #41 and #73. So, once again, #9 bumps off #41 and hands the dagger over to #73. But, what goes around, comes around.

#73 bumps off #9 and emerges the winner of this round and the whole series.

The lone survivor in friend kills friend, man kills man sordid puzzle is #73 and of course, the ubiquitous dagger.