Little explorations with HP calculators (no Prime)

[DavidM]

(12-31-2018 08:20 PM)pier4r Wrote:  Hmm, am I wrong when I think that 71 matches should be simply impossible?

Here's a straightforward way to create 21 arrangements where all starting positions (1-71) result in a "match":

1) Find all unique arrangements of a combination of 5 blue and 2 white balls:
(1, 1, 1, 1, 0, 0, 1)
(0, 1, 1, 1, 0, 1, 1)
(1, 1, 0, 0, 1, 1, 1)
(1, 1, 1, 0, 1, 1, 0)
(1, 0, 1, 0, 1, 1, 1)
(0, 1, 1, 0, 1, 1, 1)
(1, 1, 1, 1, 0, 1, 0)
(1, 1, 0, 1, 1, 0, 1)
(1, 0, 1, 1, 1, 0, 1)
(1, 1, 1, 0, 0, 1, 1)
(0, 1, 1, 1, 1, 0, 1)
(1, 1, 0, 1, 1, 1, 0)
(1, 0, 1, 1, 1, 1, 0)
(1, 0, 0, 1, 1, 1, 1)
(0, 1, 1, 1, 1, 1, 0)
(0, 0, 1, 1, 1, 1, 1)
(0, 1, 0, 1, 1, 1, 1)
(1, 1, 1, 1, 1, 0, 0)
(1, 1, 1, 0, 1, 0, 1)
(1, 1, 0, 1, 0, 1, 1)
(1, 0, 1, 1, 0, 1, 1)

2) For each of those groups, repeat the indicated pattern 20 times to form a particular arrangement of 100 1s (blue) and 40 0s (white)

Each resulting 140-ball arrangement has 71 "matches" for this problem. In other words, any contiguous subgroup of 70 elements from any of those lists will result in a group containing 50 1s and 20 0s.

These aren't the only ones, though. I believe the same logic applies if you start with a set of 10 1s and 4 0s, stepping through each unique arrangement (this time repeated only 10 times). There are undoubtedly others, but I haven't taken the time to verify them.