Dear Calendar People I said: “Each case provides an upper and lower approximation to the whole cycle at each lower complexity: 1/6 <
3/17 < 11/62 < 41/231 < 71/400 1/5 > 2/11 > 8/45 > 30/169 > 71/400 1/3 <
4/11 < 11/30 < 123/334 < 383/1040 1/2 > 3/8 > 7/19 >
130/353 > 383/1040 Here I mark the majority cycles in bold. I believe all the continued fraction convergents are included here and all the majority cycles are c.f. convergents and some of the minority cycles may also be c.f. convergents.” I have since found the continued fraction convergents. I mark them with their ordinal number in []. 1/6[2] <
3/17[4] < 11/62[6] < 41/231 < 71/400[8] 1/5[1] > 2/11[3] > 8/45 [5] > 30/169[7] > 71/400[8] 1/3[2] <
4/11[4] < 11/30 < 123/334[6] < 383/1040[8] 1/2[1] > 3/8 [3] >
7/19 [5] > 130/353[7] > 383/1040[8] Each iteration in the same column is a free iteration, which increases accuracy without increasing the structural complexity. Also as asserted every majority cycle is a convergent. Also some minority cycles are also convergents and they are the ones that have a free iteration next. Another example, which Amos’s program should be able to give all (4) levels of complexity is the 293-year cycle of 52 leap weeks ( 52/293 ): 1/6[2] <
3/17[4] < 11/62[6] < 52/293[7] 1/5[1] > 2/11[3] > 8/45 [5] > 52/293[7] Here all six of the lower level approximations are continued fraction convergents and half of them are free iterations, which is the maximum possible. Finally the truncated 33-year leap day cycle of Walter Ziobro: 1/5 <
8/33[2] < 97/400[3] 1/4[1] > 9/37 > 97/400[3] has no free iterations. Karl 16(16(17 From: Palmen,
Karl (STFC,RAL,ISIS) Dear Calendar People Here I show how cycles where the leap years are spread as smoothly as possible are composed of parts that form similar cycles of structural complexity of 1 less than the structural complexity of the whole cycle. I hope it shows some of
you the structure I see in the structural complexity. First I show it with the leap week cycle of 71 leap years in 400 years. Then with the leap month cycle of 383 leap years in 1040 years.
The cycle (71/400) has complexity 5. It divides into two parts, which as cycles have complexity 4. They are (41/231) and (30/169). The (41/231) is what one gets if one applies an ISO-like week rule to the 33-years cycle of 8 leap years. So (71/400) divides into one (41/231) and one (30/169) both of complexity 4. (41/231) divides into three (11/62) and one (8/45) both of complexity 3. (30/169) divides into two (11/62) and one (8/45). (11/62) divides into three (3/17) and one (2/11) both of complexity 2. (8/45) divides into two (3/17) and one (2/11). (3/17) divides into two (1/6) and one (1/5) both of complexity 1. (2/11) divides into one (1/6) and one (1/5). The cycle (383/1040) also has complexity 5. (383/1040) divides into two (130/353) and one (123/334) both of complexity 4. (123/334) divides into sixteen (7/19) and one (11/30) both of complexity 3. (130/353) divides into seventeen (7/19) and one (11/30). (11/30) divides into two (4/11) and one (3/8) both of complexity 2. (7/19) divides into one (4/11) and one (3/8). (4/11) divides into three (1/3) and one (1/2) both of complexity 1. (3/8) divides into two (1/3) and one (1/2). Each case provides an upper and lower approximation to the whole cycle at each lower complexity: 1/6 <
3/17 < 11/62 < 41/231 < 71/400 1/5 > 2/11 > 8/45 > 30/169 > 71/400 1/3 <
4/11 < 11/30 < 123/334 < 383/1040 1/2 > 3/8 > 7/19 >
130/353 > 383/1040 Here I mark the majority cycles in bold. I believe all the continued fraction convergents are included here and all the majority cycles are c.f. convergents and some of the minority cycles may also be c.f. convergents. For the 293-year leap week cycle of 52 leap years (complexity 4) we have: 1/6 <
3/17 < 11/62 < 52/293 1/5 > 2/11 > 8/45 > 52/293
Karl 16(16(10 |
Free forum by Nabble | Edit this page |