Thread: Cube Mountains
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Old 2008-05-30, 15:05   #9
wblipp
 
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"William"
May 2003
New Haven

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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
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