Dear calendarists!
What is a good, i.e. simple yet accurate, rule to determine when to drop a 6day week from a 366day or 61week calendar? (Alternatively, consider it a 360day, 60week calendar with a very frequent 6day leap week.) The Julian precision is rather simple to match: 365.25 = 365 + 1/4 = 366  3/4 = 364 + 1.25 = 7*52 + 7*5/28 = 7 * (52 + 5/28) = 360 + 5.25 = 6*60 + 6*7/8 = 6 * (60 + 7/8) = 6 * (61  1/8) > Drop one week every eighth year. The Gregorian precision (i.e. the targeted upper bound) is, of course, more complicated, but the resulting rule is actually simpler than the original one: 365.2425 = 365 + 97/400 = 365 + 1/4  1/100 + 1/400 = 364 + 1.2425 = 7*52 + 7*71/400 = 7 * (52 + 71/400) 360 + 5.2425 = 6*60 + 6*699/800 = 6 * (60 + 7/8  1/800) = 6 * (61  1/8  1/800) > Drop one week every eighth year except when the year number is divisible by 800. The Dee precision seems not worth it unless I'm missing a simple rule for this: 365._24 = 365 + 8/33 360 + 5._24 = 6*60 + 6*173/198 = 6 * (60 + 173/198) = 6 * (61  25/198) 
Dear Christoph and Calendar People
The Gregorian 6 * (61  1/8  1/800) is a high jitter rule, but with less jitter than 5:40:400 = 7 * (52 + 1/5  1/40 + 1/400). 6 * (61  25/198) can be a minimum jitter rule and as such, its cycle has structural complexity of at least 3, because 198 is not one different from a multiple of 25. In a minimum jitter rule, the exceptional years (with one week less) occur once every 7 or 8 years. Complexity 1: 6 * (61  1/8) = 365.25 Complexity 2: 6 * (61  2/15) = 365.2 6 * (61  3/23) = 365.217... ... 6 * (61  11/87) = 365.2414... 6 * (61  12/95) = 365.2421... 6 * (61  13/103) = 365.2427... This suggests use of a 95year cycle with 12 exceptional years, which occur 8 years apart within each 95year cycle, but 7 years apart between 95year cycles. Anything else would be more complicated, more jittery or less accurate. I can see that 33year cycle's 25/198 would arise by mixing the 12/95 with the 13/103 equally and so has complexity 3. Karl 16(15(28 Original Message From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Christoph Päper Sent: 16 November 2017 15:27 To: [hidden email] Subject: Leap rule for 366day year with 6day week Dear calendarists! What is a good, i.e. simple yet accurate, rule to determine when to drop a 6day week from a 366day or 61week calendar? (Alternatively, consider it a 360day, 60week calendar with a very frequent 6day leap week.) The Julian precision is rather simple to match: 365.25 = 365 + 1/4 = 366  3/4 = 364 + 1.25 = 7*52 + 7*5/28 = 7 * (52 + 5/28) = 360 + 5.25 = 6*60 + 6*7/8 = 6 * (60 + 7/8) = 6 * (61  1/8) > Drop one week every eighth year. The Gregorian precision (i.e. the targeted upper bound) is, of course, more complicated, but the resulting rule is actually simpler than the original one: 365.2425 = 365 + 97/400 = 365 + 1/4  1/100 + 1/400 = 364 + 1.2425 = 7*52 + 7*71/400 = 7 * (52 + 71/400) 360 + 5.2425 = 6*60 + 6*699/800 = 6 * (60 + 7/8  1/800) = 6 * (61  1/8  1/800) > Drop one week every eighth year except when the year number is divisible by 800. The Dee precision seems not worth it unless I'm missing a simple rule for this: 365._24 = 365 + 8/33 360 + 5._24 = 6*60 + 6*173/198 = 6 * (60 + 173/198) = 6 * (61  25/198) 
In reply to this post by Christoph Päper2
From: East Carolina University Calendar discussion List [[hidden email]] on behalf of Christoph Päper [[hidden email]]
Sent: Thursday, November 16, 2017 10:27 What is a good, i.e. simple yet accurate, rule to determine when to drop a 6day week from a 366day or 61week calendar? (Alternatively, consider it a 360day, 60week calendar with a very frequent 6day leap week.) The Julian precision is rather simple to match... Irv replies: "precision" is the wrong term here, you are talking about matching the mean year, which is mostly concerned with accuracy relative to some specified astronomical target, such as an equinox or solstice. In the present era the most stable astronomical mean years are those of the north solstice and the northward equinox. Precision may best be considered as reflecting the shortterm jitter of the leap cycle. The ideal leap rule for providing both accuracy and precision (minimum jitter) is a smoothly spread one, and for simplest evaluability of astronomical accuracy the leap cycle ought to be symmetrical also, which makes it valid to evaluate just the first year of each cycle relative to the desired astronomical target (typically mean equinox or solstice "events"), because the first year will always be at the average. The actual astronomical events wobble around so much that you can't match them without using very complex astronomical algorithms for your calendar, including compensating for Delta T. Your calendar could have twelve 30day months with five 6day weeks per month, for a total of 360 days in a short year, and the frequently occurring leap week could be appended to the last month or exist as a standalone minimonth at the end of the year. Positioning the leap week at the end of the year simplifies calendrical calculations because all dates in the calendar then have permanent ordinal numbers relative to the start of the year, and fixed month lengths that all contain only a whole number of weeks makes for a very simple perpetual calendar. All possible candidate leap cycles having a maximum of 1000 years per cycle are listed in the attached PDF. The instructions at the top show how to implement a smoothly spread symmetrical leap cycle. The instructions at the bottom show how to implement some higherjitter cycle alternatives that you might prefer but I don't recommend. The rows are coloured according to their suitability to the mean astronomical years of the equinoxes and solstices reckoned for the year 2020 AD. Blue is for the south solstice, with dark blue the most accurate (but keep in mind that the south solstitial year is currently rapidly getting shorter, so no calendar having a fixed mean year can approximate it well). Green is for the northward equinox, with dark green the most accurate (this mean equinoctial year will be quite stable until around the year 6000 AD). Brown is for the southward equinox, with dark brown the most accurate (but keep in mind that the southward equinoctial year is currently rapidly getting shorter, so no calendar having a fixed mean year can approximate it well). Pink is for the north solstice, with red rows indicating the most accurate (this mean solstitial year will be quite stable until around the year 13000 AD). I consider the red rows to contain the "winners", especially the 269year cycle with 34 short years per cycle because it is the shortest cycle. You may not be as impressed as I am with the longterm stability of the mean north solstitial year  please allow me to convince you by asking you to study my web page "The Length of the Seasons (On Earth)" at http://www.sym454.org/seasons/. If you are interested in any lunar associations then various columns show the number of lunar months per cycle, with those close to an integer highlighted with yellow backgrounds. I generated this list using my freeware Fixed Leap Cycle Finder spreadsheet that is on my web site. You may wish to try different settings: http://individual.utoronto.ca/kalendis/leap/index.htm#find My freeware Ford Circles spreadsheet generates a similar listing almost instantly but without the lunar information, and without the higherjitter cycles. You can find it here: http://individual.utoronto.ca/kalendis/leap/FordCirclesofLeapCycles.xls and the Ford Circles chart of 6day leap week cycles that I already have posted on my web site is here: http://individual.utoronto.ca/kalendis/leap/Ford_6d_LW.pdf.  Irv Bromberg, University of Toronto, Canada http://www.sym454.org/leap/ Candidate Calendar Cycles 6day week, year lengths 360 or 366.pdf (280K) Download Attachment 
Dear Irv sand Calendar People I have suggested the 95year cycle of 12 short years, which is listed in Irv attachment as the first brown cycle. It has a mean year of about 365.2421 days. The short years form a sequence as simple as the 33year cycle of 8 leap years and can defined symmetrically (in a Helios cycle) as the 4^{th}, 12^{th}, 20^{th}, 28^{th}, 36^{th}, 44^{th}, 52^{nd}, 60^{th}, 68^{th}, 76^{th}, 84^{th}
& 92^{nd} years of the 95year cycle. Then the 1^{st} year has an average start. All these numbers are divisible by 4, but not by 8.
The only other cycles listed in the attachment with equal simplicity are
the 103year cycle of 13 short years (mean year 365.24272 days) and
the 87year cycle of 11 short years (mean year 365.24138 days). Both these have a symmetrical cycle with the same short years, except that the 103year cycle additionally has the 100^{th} year short and the 87year
cycle does not of course have the 92^{nd} year short. At the end of the attachment, Irv listed various cycles implemented as simple cycles not necessarily spread as smoothly as possible and so may have higher jitter.
However in the case of the 87year, 95year & 103year cycle, they are spread as smoothly as possible, so do not have higher jitter. This arises because the interval (of 7) between the cycles is just 1 different from the interval (of 8) within the cycle. I have thought of a way of extending my idea of structural complexity to some cycles where the two types of year are not spread as smoothly as possible. Then
all of Irv’s simple cycles have complexity 2. The 128year cycle formed by dropping a leap year one every 128 years also has complexity 2. The Gregorian 400year cycle has complexity 3 and the Revised Julian 900year cycle has complexity 4. The ISO leap week
cycle does not have a value for this complexity, because it has 3 interval lengths. For “The Julian precision is rather simple to match...”
, I’d say “The Julian mean year is rather simple to match...”. Karl 16(16(03 From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Irv Bromberg From: East Carolina University Calendar discussion List [[hidden email]] on behalf of Christoph Päper [[hidden email]]

Dear Karll 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 On Tuesday, November 21, 2017 Karl Palmen <[hidden email]> wrote: Dear Irv sand Calendar People
I have suggested the 95year cycle of 12 short years, which is listed in Irv attachment as the first brown cycle. It has a mean year of about 365.2421 days.
The short years form a sequence as simple as the 33year cycle of 8 leap years and can defined symmetrically (in a Helios cycle) as the 4^{th}, 12^{th}, 20^{th}, 28^{th}, 36^{th}, 44^{th}, 52^{nd}, 60^{th}, 68^{th}, 76^{th}, 84^{th} & 92^{nd} years of the 95year cycle. Then the 1^{st} year has an average start. All these numbers are divisible by 4, but not by 8.
The only other cycles listed in the attachment with equal simplicity are the 103year cycle of 13 short years (mean year 365.24272 days) and the 87year cycle of 11 short years (mean year 365.24138 days). Both these have a symmetrical cycle with the same short years, except that the 103year cycle additionally has the 100^{th} year short and the 87year cycle does not of course have the 92^{nd} year short.
At the end of the attachment, Irv listed various cycles implemented as simple cycles not necessarily spread as smoothly as possible and so may have higher jitter. However in the case of the 87year, 95year & 103year cycle, they are spread as smoothly as possible, so do not have higher jitter. This arises because the interval (of 7) between the cycles is just 1 different from the interval (of 8) within the cycle.
I have thought of a way of extending my idea of structural complexity to some cycles where the two types of year are not spread as smoothly as possible. Then all of Irv’s simple cycles have complexity 2. The 128year cycle formed by dropping a leap year one every 128 years also has complexity 2. The Gregorian 400year cycle has complexity 3 and the Revised Julian 900year cycle has complexity 4. The ISO leap week cycle does not have a value for this complexity, because it has 3 interval lengths.
For “The Julian precision is rather simple to match...” , I’d say “The Julian mean year is rather simple to match...”.
Karl
16(16(03
From: East Carolina University Calendar discussion List [mailto:CALNDRL@...] On Behalf Of Irv Bromberg Sent: 16 November 2017 18:42 To: CALNDRL@... Subject: Re: Leap rule for 366day year with 6day week
From: East Carolina University Calendar discussion List [CALNDRL@...] on behalf of Christoph Päper [christoph.paeper@...]

In reply to this post by Christoph Päper2
Dear Walter and Calendar People The 128year 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 (20170519) 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 33year cycle has complexity 2. 97/400
> 12/97 > 1/12 > 12. The 400year cycle formed from 12 33year 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 334year leap month cycle has complexity
4. 11/62 > 7/11 (4/11) > ¾ (¼) > 4. The 62year leap week cycle has complexity
3. 52/293
> 33/52 (19/52) > 14/19 (5/19) > 4/5 (1/5) > 5. The 293year leap week cycle has complexity
4. 71/400
> 45/71 (26/71) > 19/26 (7/26) > 5/7 (2/7) > ½
> 2. The 400year leap week cycle has complexity 5. 83/95 (12/95)
> 11/12 (1/12) > 12. The 95year cycle 6day 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 400year cycle has
12 33year cycles one of which is extended by an Olympiad and the 334year cycle has
17 Metonic cycles one of which is truncated by an Octaeteris. Karl 16(16(06 From: Walter J Ziobro [mailto:[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 On Tuesday, November 21, 2017 Karl Palmen <[hidden email]>
wrote: Dear Irv sand Calendar People I have suggested the 95year cycle of 12 short years, which is listed in Irv attachment as the first brown cycle.
It has a mean year of about 365.2421 days. The short years form a sequence as simple as the 33year cycle of 8 leap years and can defined symmetrically (in a
Helios cycle) as the 4^{th}, 12^{th}, 20^{th}, 28^{th}, 36^{th}, 44^{th}, 52^{nd},
60^{th}, 68^{th}, 76^{th}, 84^{th} & 92^{nd} years of the 95year cycle. Then the 1^{st} year has an average start. All these numbers are divisible by 4, but not by 8.
The only other cycles listed in the attachment with equal simplicity are
the 103year cycle of 13 short years (mean year 365.24272 days) and
the 87year cycle of 11 short years (mean year 365.24138 days). Both these have a symmetrical cycle with the same short years, except that the 103year cycle additionally
has the 100^{th} year short and the 87year cycle does not of course have the 92^{nd} year short. At the end of the attachment, Irv listed various cycles implemented as simple cycles not necessarily
spread as smoothly as possible and so may have higher jitter. However in the case of the 87year, 95year & 103year cycle, they are spread as smoothly as possible, so do not have higher jitter. This arises because the interval (of 7) between the cycles is
just 1 different from the interval (of 8) within the cycle. I have thought of a way of extending my idea of structural complexity to some cycles where the two
types of year are not spread as smoothly as possible. Then all of Irv’s simple cycles have complexity 2. The 128year cycle formed by dropping a leap year one every 128 years also has complexity 2. The Gregorian 400year cycle has complexity 3 and the Revised
Julian 900year cycle has complexity 4. The ISO leap week cycle does not have a value for this complexity, because it has 3 interval lengths. For “The Julian precision is rather simple to match...”
, I’d say “The Julian mean year is rather simple to match...”. Karl 16(16(03 From: East Carolina University Calendar discussion List [[hidden email]]
On Behalf Of Irv Bromberg From: East Carolina University Calendar discussion List [CALNDR[hidden email]]
on behalf of Christoph Päper [[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 128year 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 (20170519) 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 33year cycle has complexity 2.
97/400 > 12/97 > 1/12 > 12. The 400year cycle formed from 12 33year 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 334year leap month cycle has complexity 4.
11/62 > 7/11 (4/11) > ¾ (¼) > 4. The 62year leap week cycle has complexity 3.
52/293 > 33/52 (19/52) > 14/19 (5/19) > 4/5 (1/5) > 5. The 293year leap week cycle has complexity 4.
71/400 > 45/71 (26/71) > 19/26 (7/26) > 5/7 (2/7) > ½ > 2. The 400year leap week cycle has complexity 5.
83/95 (12/95) > 11/12 (1/12) > 12. The 95year cycle 6day 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 400year cycle has 12 33year cycles one of which is extended by an Olympiad and the 334year cycle has 17 Metonic cycles one of which is truncated by an Octaeteris.
Karl
16(16(06
From: Walter J Ziobro [mailto:walterziobro@...] Sent: 24 November 2017 03:47 To: Palmen, Karl (STFC,RAL,ISIS); CALNDRL@... Subject: Re: Leap rule for 366day year with 6day week
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
On Tuesday, November 21, 2017 Karl Palmen <[hidden email]> wrote: Dear Irv sand Calendar People
I have suggested the 95year cycle of 12 short years, which is listed in Irv attachment as the first brown cycle. It has a mean year of about 365.2421 days.
The short years form a sequence as simple as the 33year cycle of 8 leap years and can defined symmetrically (in a Helios cycle) as the 4^{th}, 12^{th}, 20^{th}, 28^{th}, 36^{th}, 44^{th}, 52^{nd}, 60^{th}, 68^{th}, 76^{th}, 84^{th} & 92^{nd} years of the 95year cycle. Then the 1^{st} year has an average start. All these numbers are divisible by 4, but not by 8.
The only other cycles listed in the attachment with equal simplicity are the 103year cycle of 13 short years (mean year 365.24272 days) and the 87year cycle of 11 short years (mean year 365.24138 days). Both these have a symmetrical cycle with the same short years, except that the 103year cycle additionally has the 100^{th} year short and the 87year cycle does not of course have the 92^{nd} year short.
At the end of the attachment, Irv listed various cycles implemented as simple cycles not necessarily spread as smoothly as possible and so may have higher jitter. However in the case of the 87year, 95year & 103year cycle, they are spread as smoothly as possible, so do not have higher jitter. This arises because the interval (of 7) between the cycles is just 1 different from the interval (of 8) within the cycle.
I have thought of a way of extending my idea of structural complexity to some cycles where the two types of year are not spread as smoothly as possible. Then all of Irv’s simple cycles have complexity 2. The 128year cycle formed by dropping a leap year one every 128 years also has complexity 2. The Gregorian 400year cycle has complexity 3 and the Revised Julian 900year cycle has complexity 4. The ISO leap week cycle does not have a value for this complexity, because it has 3 interval lengths.
For “The Julian precision is rather simple to match...” , I’d say “The Julian mean year is rather simple to match...”.
Karl
16(16(03
From: East Carolina University Calendar discussion List [[hidden email]]
On Behalf Of Irv Bromberg
From: East Carolina University Calendar discussion List [CALNDR[hidden email]]
on behalf of Christoph Päper [[hidden email]]

Dear Walter and Calendar People The rulebased 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 128year cycles as described followed by 4 33year 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 2820year 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 128year cycle it contains. Karl 16(16(06 From: Walter J Ziobro [mailto:[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 128year 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 (20170519) 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 33year cycle has complexity 2. 97/400
> 12/97 > 1/12 > 12. The 400year cycle formed from 12 33year 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 334year leap month cycle has complexity
4. 11/62 > 7/11 (4/11) > ¾ (¼) > 4. The 62year leap week cycle has complexity
3. 52/293
> 33/52 (19/52) > 14/19 (5/19) > 4/5 (1/5) > 5. The 293year leap week cycle has complexity
4. 71/400
> 45/71 (26/71) > 19/26 (7/26) > 5/7 (2/7) > ½
> 2. The 400year leap week cycle has complexity 5. 83/95 (12/95)
> 11/12 (1/12) > 12. The 95year cycle 6day 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 400year cycle has 12 33year cycles one of which is extended by an Olympiad and the 334year 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 On Tuesday, November 21, 2017 Karl Palmen <[hidden email]>
wrote: Dear Irv sand Calendar People I have suggested the 95year cycle of 12 short years, which is listed in Irv attachment as the first brown cycle.
It has a mean year of about 365.2421 days. The short years form a sequence as simple as the 33year cycle of 8 leap years and can defined symmetrically (in a
Helios cycle) as the 4^{th}, 12^{th}, 20^{th}, 28^{th}, 36^{th}, 44^{th}, 52^{nd},
60^{th}, 68^{th}, 76^{th}, 84^{th} & 92^{nd} years of the 95year cycle. Then the 1^{st} year has an average start. All these numbers are divisible by 4, but not by 8.
The only other cycles listed in the attachment with equal simplicity are
the 103year cycle of 13 short years (mean year 365.24272 days) and
the 87year cycle of 11 short years (mean year 365.24138 days). Both these have a symmetrical cycle with the same short years, except that the 103year cycle additionally
has the 100^{th} year short and the 87year cycle does not of course have the 92^{nd} year short. At the end of the attachment, Irv listed various cycles implemented as simple cycles not necessarily
spread as smoothly as possible and so may have higher jitter. However in the case of the 87year, 95year & 103year cycle, they are spread as smoothly as possible, so do not have higher jitter. This arises because the interval (of 7) between the cycles is
just 1 different from the interval (of 8) within the cycle. I have thought of a way of extending my idea of structural complexity to some cycles where the two
types of year are not spread as smoothly as possible. Then all of Irv’s simple cycles have complexity 2. The 128year cycle formed by dropping a leap year one every 128 years also has complexity 2. The Gregorian 400year cycle has complexity 3 and the Revised
Julian 900year cycle has complexity 4. The ISO leap week cycle does not have a value for this complexity, because it has 3 interval lengths. For “The Julian precision is rather simple to match...”
, I’d say “The Julian mean year is rather simple to match...”. Karl 16(16(03 From: East
Carolina University Calendar discussion List [[hidden email]]
On Behalf Of Irv Bromberg From: East Carolina University Calendar discussion List [CALNDR[hidden email]]
on behalf of Christoph Päper [[hidden email]]

Dear Karl That is the algorithm proposed by Ahmad Birashk It differs from the 33333329 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 rulebased 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 128year cycles as described followed by 4 33year 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 2820year 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 128year cycle it contains.
Karl
16(16(06
From: Walter J Ziobro [mailto:walterziobro@...] Sent: 24 November 2017 15:46 To: Palmen, Karl (STFC,RAL,ISIS); CALNDRL@... Subject: RE: Leap rule for 366day year with 6day week
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 128year 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 (20170519) 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 33year cycle has complexity 2.
97/400 > 12/97 > 1/12 > 12. The 400year cycle formed from 12 33year 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 334year leap month cycle has complexity 4.
11/62 > 7/11 (4/11) > ¾ (¼) > 4. The 62year leap week cycle has complexity 3.
52/293 > 33/52 (19/52) > 14/19 (5/19) > 4/5 (1/5) > 5. The 293year leap week cycle has complexity 4.
71/400 > 45/71 (26/71) > 19/26 (7/26) > 5/7 (2/7) > ½ > 2. The 400year leap week cycle has complexity 5.
83/95 (12/95) > 11/12 (1/12) > 12. The 95year cycle 6day 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 400year cycle has 12 33year cycles one of which is extended by an Olympiad and the 334year 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
On Tuesday, November 21, 2017 Karl Palmen <[hidden email]> wrote: Dear Irv sand Calendar People
I have suggested the 95year cycle of 12 short years, which is listed in Irv attachment as the first brown cycle. It has a mean year of about 365.2421 days.
The short years form a sequence as simple as the 33year cycle of 8 leap years and can defined symmetrically (in a Helios cycle) as the 4^{th}, 12^{th}, 20^{th}, 28^{th}, 36^{th}, 44^{th}, 52^{nd}, 60^{th}, 68^{th}, 76^{th}, 84^{th} & 92^{nd} years of the 95year cycle. Then the 1^{st} year has an average start. All these numbers are divisible by 4, but not by 8.
The only other cycles listed in the attachment with equal simplicity are the 103year cycle of 13 short years (mean year 365.24272 days) and the 87year cycle of 11 short years (mean year 365.24138 days). Both these have a symmetrical cycle with the same short years, except that the 103year cycle additionally has the 100^{th} year short and the 87year cycle does not of course have the 92^{nd} year short.
At the end of the attachment, Irv listed various cycles implemented as simple cycles not necessarily spread as smoothly as possible and so may have higher jitter. However in the case of the 87year, 95year & 103year cycle, they are spread as smoothly as possible, so do not have higher jitter. This arises because the interval (of 7) between the cycles is just 1 different from the interval (of 8) within the cycle.
I have thought of a way of extending my idea of structural complexity to some cycles where the two types of year are not spread as smoothly as possible. Then all of Irv’s simple cycles have complexity 2. The 128year cycle formed by dropping a leap year one every 128 years also has complexity 2. The Gregorian 400year cycle has complexity 3 and the Revised Julian 900year cycle has complexity 4. The ISO leap week cycle does not have a value for this complexity, because it has 3 interval lengths.
For “The Julian precision is rather simple to match...” , I’d say “The Julian mean year is rather simple to match...”.
Karl
16(16(03
From: East
Carolina University Calendar discussion List [[hidden email]]
On Behalf Of Irv Bromberg
From: East Carolina University Calendar discussion List [CALNDR[hidden email]]
on behalf of Christoph Päper [[hidden email]]

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