Dear Walter and Calendar People 683/2820 = 0.24219858… 31/128 = 0.2421875 So the difference is 0.00001108… , which is 1 day in 90240 years , 1 hour in 3760 years, 1 minute in 62 2/3 years, and so almost one second per year. If the 132-year cycles were to occur once every 21^{st} cycle. The difference would be 1 day in 86144 years, which is just over 1 second per year. So
if an approximation to 365 days 5h 48m 46s were desired this would be a better choice. Also if the leap years were spread as smoothly as possible the cycle would break up into four 673-year cycles of 163 leap years
(5*128+33=673 years, 5*31+8 = 163 leap years).
I expect 2820 years was chosen, because it is a round number divisible by 60. Karl 16(16(09 From: Walter J Ziobro [mailto:[hidden email]]
Dear Karl That is the algorithm proposed by Ahmad Birashk It differs from the 33-33-33-29 cycle of 128 years by having a 132 year cycle every 22nd cycle Does that 132 year cycle make a significant difference in the long run? Walter Ziobro Sent from AOL Mobile Mail On Friday, November 24, 2017 karl.palmen <[hidden email]>
wrote: Dear Walter and Calendar People The rule-based Iranian calendar does not have its leap years spread as smoothly as possible. It has
2820 years of which 683 are leap years. If the leap years were spread as smoothly as possible, we’d get 683/2820
-> 88/683 -> 67/88 (21/88) -> 4/21 -> ¼ -> 4. And so it would have complexity
5. The actual cycle consists of 21 128-year cycles as described followed by 4 33-year cycles. The interval
cycle of the interval cycle thus has 21 (8,8,8,7)s followed by one (8,8,8,8) and so the interval cycle of this is 20 fours and 1 eight, which has complexity 1 in my extended definition of complexity and so the actual 2820-year cycle has a complexity of
4 in my extended definition, one less than the 5 that would occur is the leap years were spread as smoothly as possible and one more than the 128-year cycle it contains. Karl 16(16(06 From: Walter J Ziobro [[hidden email]]
Dear Karl I thank you for that information I am becoming more intrigued by this cycle. It appears to be the rule based cycle that most nearly follows the astronomical cycles of the Iranian and solar Chinese calendars
Walter Ziobro Sent from AOL Mobile Mail On Friday, November 24, 2017 karl.palmen <[hidden email]>
wrote: Dear Walter and Calendar People The 128-year cycle as described has complexity 3. 0/1 [1] ¼ [2] 8/33 [3] 31/128. Each pair of consecutive fractions listed here has their Ford circles
touching and this a shortest such path. https://en.wikipedia.org/wiki/Ford_circle
I’ve found an algorithm to calculate the complexity of a cycle. The idea of the intervals cycles
of interval cycles (2017-05-19) gave rise to it. Take the cycle as the number of long years divided by the number of years: 31/128 If this fraction is greater than ½, subtract it from 1 and then use that fraction. 31/128 Then invert the fraction an take the fractional part of the result 128/31
-> 4/31 This is one step or iteration.
Repeat till one gets an integer after the inversion. 31/128
-> 4/31 -> ¾ (¼) -> 4. Three steps, therefore the complexity is three. Here are some other examples:
8/33
-> 1/8 -> 8. The 33-year cycle has complexity 2. 97/400
-> 12/97 -> 1/12 -> 12. The 400-year cycle formed from 12 33-year cycles and one Olympiad has complexity
3. 7/19
-> 5/7 (2/7) -> ½ -> 2. The Metonic leap month cycle has complexity
3. 123/334
-> 88/123 (35/123) -> 18/35 (17/35) -> 1/17 -> 17. The 334-year leap month cycle has complexity
4. 11/62 -> 7/11 (4/11) -> ¾ (¼) -> 4. The 62-year leap week cycle has complexity
3. 52/293
-> 33/52 (19/52) -> 14/19 (5/19) -> 4/5 (1/5) -> 5. The 293-year leap week cycle has complexity
4. 71/400
-> 45/71 (26/71) -> 19/26 (7/26) -> 5/7 (2/7) -> ½
-> 2. The 400-year leap week cycle has complexity 5. 83/95 (12/95)
-> 11/12 (1/12) -> 12. The 95-year cycle 6-day week cycle has complexity
2. This method is the same as counting the steps of the continued fraction, except for the subtraction
from 1 if greater than a half. I then realise, it would be equivalent to the continued fraction method, but with rounding to nearest integer instead of rounding down, except that the sign is ignored. The final integer is equal to the number of parts of the cycles, which as cycles have complexity
1 less. For example the 400-year cycle has 12 33-year cycles one of which is extended by an Olympiad and the 334-year cycle has
17 Metonic cycles one of which is truncated by an Octaeteris. Karl 16(16(06 From: Walter
J Ziobro [[hidden email]]
Dear Karl What is the complexity level of a leap year cycle that has 3 33 year cycles plus 1 29 year cycle, for a total of 128 years? Walter Ziobro Sent from AOL Mobile Mail |
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