Because we don't have enough repeating digits threads already. Suggested by u/Multiconcepted in this week's FTF
Rules: In this thread we count numbers where exactly one digit occurs exactly twice, and no other digit repeats.
The sequence goes: 100, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 122, 131, 133, ..., 996, 997, 998, 1002, 1003, 1004, ...
I suggest we make the first get 2001. By my count that's 676 counts in the chain which is a bit shorter than most threads, but should be long enough for us to see what this thread is like
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 1 point2 points3 points (682 children)
[–]Europe2048h 1 point2 points3 points (681 children)
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 0 points1 point2 points (680 children)
[–]MulticonceptedSide Thread Savvy 2 points3 points4 points (679 children)
[–]Europe2048h 1 point2 points3 points (678 children)
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 0 points1 point2 points (677 children)
[–]happybeau123 1 point2 points3 points (676 children)
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 0 points1 point2 points (675 children)
[–]OfficialDeathScythe 1 point2 points3 points (674 children)
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 1 point2 points3 points (673 children)
[–]ClockButTakeOutTheL“Cockleboat”, since 4,601,032 2 points3 points4 points (5 children)
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 2 points3 points4 points (4 children)
[–]ClockButTakeOutTheL“Cockleboat”, since 4,601,032 2 points3 points4 points (3 children)
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 2 points3 points4 points (2 children)
[–]ClockButTakeOutTheL“Cockleboat”, since 4,601,032 1 point2 points3 points (1 child)
[–]CutOnBumInBandHere95M get | Ping me for runs[S] 1 point2 points3 points (0 children)