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Wow. Just discovered I meisread 15th Dec puzzle.

I can tell you that the number of combinations of n As and Bs which contain at at least one uninterrupted sequence of 3 As is 2^n - F3(n+3) where F3 is the fibonaccia variant adding 3 numbers (1,1,2,4,7,13,24 etc.).
Only took me about 8 hours (with some small help form OEIS for the 2 As problem)
Neal (@zbvif)
on /blog/46
               

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