Evo i rastava na cikluse pa mozda netko uoci pravilnost:
[code:1]<< DiscreteMath`Combinatorica`
Table[x = {}; Table[If[OddQ[i], x = Append[x, i], x = Prepend[x, i]], {i, 1, n}]; Print[n, ": ", StringReplace[(StringTake[#, {2, StringLength[#]-1}] &)[ToString[ToCycles[x]]], {"{" ->"(", "}," -> ")", "}" -> ")", " " -> ""}]], {n, 1, 32}][/code:1]
[quote="Mathematica"]1: (1)
2: (2,1)
3: (2,1)(3)
4: (4,3,1)(2)
5: (4,3,1)(2)(5)
6: (6,5,3,2,4,1)
7: (6,5,3,2,4,1)(7)
8: (8,7,5,1)(6,3,4,2)
9: (8,7,5,1)(6,3,4,2)(9)
10: (10,9,7,3,6,1)(8,5,2)(4)
11: (10,9,7,3,6,1)(8,5,2)(4)(11)
12: (12,11,9,5,4,6,2,10,7,1)(8,3)
13: (12,11,9,5,4,6,2,10,7,1)(8,3)(13)
14: (14,13,11,7,2,12,9,3,10,5,6,4,8,1)
15: (14,13,11,7,2,12,9,3,10,5,6,4,8,1)(15)
16: (16,15,13,9,1)(14,11,5,8,2)(12,7,4,10,3)(6)
17: (16,15,13,9,1)(14,11,5,8,2)(12,7,4,10,3)(6)(17)
18: (18,17,15,11,3,14,9,2,16,13,7,6,8,4,12,5,10,1)
19: (18,17,15,11,3,14,9,2,16,13,7,6,8,4,12,5,10,1)(19)
20: (20,19,17,13,5,12,3,16,11,1)(18,15,9,4,14,7,8,6,10,2)
21: (20,19,17,13,5,12,3,16,11,1)(18,15,9,4,14,7,8,6,10,2)(21)
22: (22,21,19,15,7,10,4,16,9,6,12,1)(20,17,11,2)(18,13,3)(14,5)(8)
23: (22,21,19,15,7,10,4,16,9,6,12,1)(20,17,11,2)(18,13,3)(14,5)(8)(23)
24: (24,23,21,17,9,8,10,6,14,3,20,15,5,16,7,12,2,22,19,13,1)(18,11,4)
25: (24,23,21,17,9,8,10,6,14,3,20,15,5,16,7,12,2,22,19,13,1)(18,11,4)(25)
26: (26,25,23,19,11,6,16,5,18,9,10,8,12,4,20,13,2,24,21,15,3,22,17,7,14,1)
27: (26,25,23,19,11,6,16,5,18,9,10,8,12,4,20,13,2,24,21,15,3,22,17,7,14,1)(27)
28: (28,27,25,21,13,4,22,15,1)(26,23,17,5,20,11,8,14,2)(24,19,9,12,6,18,7,16,3)(10)
29: (28,27,25,21,13,4,22,15,1)(26,23,17,5,20,11,8,14,2)(24,19,9,12,6,18,7,16,3)(10)(29)
30: (30,29,27,23,15,2,28,25,19,7,18,5,22,13,6,20,9,14,4,24,17,3,26,21,11,10,12,8,16,1)
31: (30,29,27,23,15,2,28,25,19,7,18,5,22,13,6,20,9,14,4,24,17,3,26,21,11,10,12,8,16,1)(31)
32: (32,31,29,25,17,1)(30,27,21,9,16,2)(28,23,13,8,18,3)(26,19,5,24,15,4)(22,11,12,10,14,6)(20,7)[/quote]
Fasciniraju me ovi koji su u jednom velikom ciklusu...
Evo i rastava na cikluse pa mozda netko uoci pravilnost:
Kod: | << DiscreteMath`Combinatorica`
Table[x = {}; Table[If[OddQ[i], x = Append[x, i], x = Prepend[x, i]], {i, 1, n}]; Print[n, ": ", StringReplace[(StringTake[#, {2, StringLength[#]-1}] &)[ToString[ToCycles[x]]], {"{" ->"(", "}," -> ")", "}" -> ")", " " -> ""}]], {n, 1, 32}] |
Mathematica (napisa): | 1: (1)
2: (2,1)
3: (2,1)(3)
4: (4,3,1)(2)
5: (4,3,1)(2)(5)
6: (6,5,3,2,4,1)
7: (6,5,3,2,4,1)(7)
8: (8,7,5,1)(6,3,4,2)
9: (8,7,5,1)(6,3,4,2)(9)
10: (10,9,7,3,6,1)(8,5,2)(4)
11: (10,9,7,3,6,1)(8,5,2)(4)(11)
12: (12,11,9,5,4,6,2,10,7,1)(8,3)
13: (12,11,9,5,4,6,2,10,7,1)(8,3)(13)
14: (14,13,11,7,2,12,9,3,10,5,6,4,8,1)
15: (14,13,11,7,2,12,9,3,10,5,6,4,8,1)(15)
16: (16,15,13,9,1)(14,11,5,8,2)(12,7,4,10,3)(6)
17: (16,15,13,9,1)(14,11,5,8,2)(12,7,4,10,3)(6)(17)
18: (18,17,15,11,3,14,9,2,16,13,7,6,8,4,12,5,10,1)
19: (18,17,15,11,3,14,9,2,16,13,7,6,8,4,12,5,10,1)(19)
20: (20,19,17,13,5,12,3,16,11,1)(18,15,9,4,14,7,8,6,10,2)
21: (20,19,17,13,5,12,3,16,11,1)(18,15,9,4,14,7,8,6,10,2)(21)
22: (22,21,19,15,7,10,4,16,9,6,12,1)(20,17,11,2)(18,13,3)(14,5)(8)
23: (22,21,19,15,7,10,4,16,9,6,12,1)(20,17,11,2)(18,13,3)(14,5)(8)(23)
24: (24,23,21,17,9,8,10,6,14,3,20,15,5,16,7,12,2,22,19,13,1)(18,11,4)
25: (24,23,21,17,9,8,10,6,14,3,20,15,5,16,7,12,2,22,19,13,1)(18,11,4)(25)
26: (26,25,23,19,11,6,16,5,18,9,10,8,12,4,20,13,2,24,21,15,3,22,17,7,14,1)
27: (26,25,23,19,11,6,16,5,18,9,10,8,12,4,20,13,2,24,21,15,3,22,17,7,14,1)(27)
28: (28,27,25,21,13,4,22,15,1)(26,23,17,5,20,11,8,14,2)(24,19,9,12,6,18,7,16,3)(10)
29: (28,27,25,21,13,4,22,15,1)(26,23,17,5,20,11,8,14,2)(24,19,9,12,6,18,7,16,3)(10)(29)
30: (30,29,27,23,15,2,28,25,19,7,18,5,22,13,6,20,9,14,4,24,17,3,26,21,11,10,12,8,16,1)
31: (30,29,27,23,15,2,28,25,19,7,18,5,22,13,6,20,9,14,4,24,17,3,26,21,11,10,12,8,16,1)(31)
32: (32,31,29,25,17,1)(30,27,21,9,16,2)(28,23,13,8,18,3)(26,19,5,24,15,4)(22,11,12,10,14,6)(20,7) |
Fasciniraju me ovi koji su u jednom velikom ciklusu...
_________________ Vedran Krcadinac
Ljudi su razliciti, a nula je paran broj.
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