Thread: Cube Mountains View Single Post
 2008-05-30, 15:05 #9 wblipp     "William" May 2003 New Haven 3×787 Posts For odd n It's easier to count the symmetric mountains than I has thought. For 180 degrees, you build a mountain on half of the non-central squares plus the central square. When you rotate that to get a symmetric mountain on the full base, the central square rotates onto itself, but that's OK. The same process works for counting 90 degree symmetric mountains. Hence the number of unique mountains on the 3x3 base is (M(9)-M(5))/4 + (M(5)-M(3))/2 + M(3) = 1771957 And the 5x5 base has (M(25)-M(13))/4 + (M(13)-M(7))/2 + M(7) = 26674341359618944088110485277 There's a substantial chance I made an error in calculating M(25). Here are my calculations to check against: 0 1 1 1 2 3 3 13 4 75 5 541 6 4683 7 47293 8 545835 9 7087261 10 102247563 11 1622632573 12 28091567595 13 526858348381 14 10641342970443 15 230283190977853 16 5315654681981355 17 130370767029135901 18 3385534663256845323 19 92801587319328411133 20 2677687796244384203115 21 81124824998504073881821 22 2574844419803190384544203 23 85438451336745709294580413 24 2958279121074145472650648875 25 106697365438475775825583498141 Last fiddled with by wblipp on 2008-05-30 at 15:16 Reason: Fixed a calculation error