Dear Walter and Calendar People Reply below From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Walter J Ziobro Dear Karl and Calendar List: KARL REPLIES: Walter’s idea runs into problems after the 33^{rd} month of each 34 or 33 is the leap month. What happens after then? One solution is to jump over the 34^{th} to the 1^{st} to smoothly drop a leap month. Then 33 of these postponements would be needed in a 6840year
cycle, 30 after 11 Metonic cycles and 3 after 3 Metonic cycles so following a subcycle of 120 Metonic cycles of 2280 years. Another idea is to have exactly 30 permitted positions for the leap month in each 34 and 33, then postponement to the next permitted position would occur exactly
once every 12 Metonic cycles of 228 years. Whenever the 30^{th}, 31^{st} or 32^{nd} month in each 33 & 34 were a leap month, the leap month years would follow the Hebrew 19year
cycle. No such cycle would create the smoothest possible distribution of leap months, because some leap months will occur at least 35 months after the previous, whenever
a postponement occurs. I expect it would add a day or two to the solar jitter of about one month. If the leap months were spread as smoothly as possible, we’d have 168month periods of 3433343334 in addition to the 235month periods of 34333433343334
all with fixed leap month. Karl 16(01(19
Original Message Dear Irv & Calendar People I’ve already mentioned an algorithm for cutting calendar cycles where the leap years are spread as smoothly as possible into symmetrical
parts, grouped in larger symmetrical parts. Now I show a similar algorithm where the parts have the leap years as late as possible. In the first iteration cut after any leap year followed by a common year. For example: Hebrew 19year cycle: ccLccLcLccLccLccLcL
, Symmetrical 17year leap week cycle:
ccLcccccLcccccLcc
,
33year cycle: cccLcccLcccLcccLcccLcccLcccLcccLc, Julian 4year cycle:
cccL
Then there should either be two different types of part (joining parts over end of cycle) or just one part in which case the algorithm
has finished. The part with the
greatest proportion of leap years and so shortest if leap years are in minority (shown in
red) is designated as leap year for
the next iteration, even if such parts are in a majority. The examples then become: ccLccLcLccLccLccLcL
, ccLcccccLcccccLcc
,
cccLcccLcccLcccLcccLcccLcccLcccLc The second example has only one double cut and so the algorithm has
finished. If the cycle is started
after the double cut (at 4^{th} year) the leap years will be as late as possible and if the year before the double cut (3^{rd} year) is year Z as defined in earlier notes and if designated year 0, makes K=0.
The third example also has only one double cut and so year Z is the 32^{nd} year and if started on the 33^{rd} year,
the leap years would occurs as late as possible. The first example, which corresponds to the Hebrew 19year cycle, has a third iteration: ccLccLcLccLccLccLcL It has only one triple cut, which is between the 8^{th} and 9^{th} year. So the 8^{th} year is year Z and the
9^{th} year is the latest starting year and has subsequent leap years as late as possible.
I may look into Irv’s 353year cycle with 130 leap years and K=269 in a later note. I’ll say at present that the second iteration cuts
it into 11s and 8s. I expect the final cut to occur in the 4^{th} iteration after the 169^{th} year, which I reckoned to be year Z (year with 0 remainder). For leap week calendars of sufficient accuracy, the first iteration cuts into 6s & 5s, the second iteration into 17s and 11s and the
third iteration cuts into 62s and either 45s or 79s, except of course the 62year cycle. The number of iterations that this algorithm runs may be considered to be a measure of the structural complexity of the cycle.
The is also a similar mirror image algorithm, which cuts into parts that have the leap years as early as possible. Karl 16(01(13

Dear Karl I thank you for that info I like your second suggestion, that there be 30 leap month shifts, one every 228 years This can be accomplished by skipping a designated leap month every 10 cycles, such that the designated leap month s in the 3334 month groups would be 110, 1221, and 2332 Walter Ziobro Sent from AOL Mobile Mail On Tuesday, September 20, 2016 Karl Palmen <[hidden email]> wrote: Dear Walter and Calendar People
Reply below
From: East Carolina University Calendar discussion List [mailto:CALNDRL@...]
On Behalf Of Walter J Ziobro
Dear Karl and Calendar List:
KARL REPLIES: Walter’s idea runs into problems after the 33^{rd} month of each 34 or 33 is the leap month. What happens after then? One solution is to jump over the 34^{th} to the 1^{st} to smoothly drop a leap month. Then 33 of these postponements would be needed in a 6840year cycle, 30 after 11 Metonic cycles and 3 after 3 Metonic cycles so following a subcycle of 120 Metonic cycles of 2280 years.
Another idea is to have exactly 30 permitted positions for the leap month in each 34 and 33, then postponement to the next permitted position would occur exactly once every 12 Metonic cycles of 228 years.
Whenever the 30^{th}, 31^{st} or 32^{nd} month in each 33 & 34 were a leap month, the leap month years would follow the Hebrew 19year cycle.
No such cycle would create the smoothest possible distribution of leap months, because some leap months will occur at least 35 months after the previous, whenever a postponement occurs. I expect it would add a day or two to the solar jitter of about one month. If the leap months were spread as smoothly as possible, we’d have 168month periods of 3433343334 in addition to the 235month periods of 34333433343334 all with fixed leap month.
Karl
16(01(19
Original Message Dear Irv & Calendar People
I’ve already mentioned an algorithm for cutting calendar cycles where the leap years are spread as smoothly as possible into symmetrical parts, grouped in larger symmetrical parts.
Now I show a similar algorithm where the parts have the leap years as late as possible.
In the first iteration cut after any leap year followed by a common year. For example:
Hebrew 19year cycle: ccLccLcLccLccLccLcL , Symmetrical 17year leap week cycle: ccLcccccLcccccLcc , 33year cycle: cccLcccLcccLcccLcccLcccLcccLcccLc, Julian 4year cycle: cccL
Then there should either be two different types of part (joining parts over end of cycle) or just one part in which case the algorithm has finished. The part with the greatest proportion of leap years and so shortest if leap years are in minority (shown in red) is designated as leap year for the next iteration, even if such parts are in a majority. The examples then become:
ccLccLcLccLccLccLcL , ccLcccccLcccccLcc , cccLcccLcccLcccLcccLcccLcccLcccLc
The second example has only one double cut and so the algorithm has finished. If the cycle is started after the double cut (at 4^{th} year) the leap years will be as late as possible and if the year before the double cut (3^{rd} year) is year Z as defined in earlier notes and if designated year 0, makes K=0. The third example also has only one double cut and so year Z is the 32^{nd} year and if started on the 33^{rd} year, the leap years would occurs as late as possible.
The first example, which corresponds to the Hebrew 19year cycle, has a third iteration:
ccLccLcLccLccLccLcL
It has only one triple cut, which is between the 8^{th} and 9^{th} year. So the 8^{th} year is year Z and the 9^{th} year is the latest starting year and has subsequent leap years as late as possible.
I may look into Irv’s 353year cycle with 130 leap years and K=269 in a later note. I’ll say at present that the second iteration cuts it into 11s and 8s. I expect the final cut to occur in the 4^{th} iteration after the 169^{th} year, which I reckoned to be year Z (year with 0 remainder).
For leap week calendars of sufficient accuracy, the first iteration cuts into 6s & 5s, the second iteration into 17s and 11s and the third iteration cuts into 62s and either 45s or 79s, except of course the 62year cycle.
The number of iterations that this algorithm runs may be considered to be a measure of the structural complexity of the cycle.
The is also a similar mirror image algorithm, which cuts into parts that have the leap years as early as possible.
Karl
16(01(13 
Dear Walter and Calendar People Thank you Walter for your reply. I don’t understand what Walter means after “This can be accomplished by skipping”. It needs explaining in more detail. I had suggested the 30 months of each 34 or 33 are permitted leap months as so 4 or 3 months of each 34 or 33 respectively are forbidden months.
These could be (5 15 24 34) in a 34 and (10 19 29) in a 33 forming a Helios cycle over each 34+33, which is interrupted, between 19year cycles when there are
two consecutive 34s. This could be simplified to (4 14 24 34) in a 34 and (9 19 29) in a 33. It could be simplified further to (4 14 24 34) in a 34 and (4 14 24) in a 33.
This further simplification ensures that for any 228year cycle, the leap months occur in the same position (counting from start) in both the 34s and the 33s,
whereas my earlier suggestions would for some 228year cycles, have the leap months occurring 1 month earlier in a 33 than in a 34, assuming they initially occurred at the first month of both. In this simplest suggestion a month would be skipped every
nine 228year cycles within a 6840year cycle starting with the 4^{th} (from 685^{th} year), then the 13^{th} (from 2737^{th} year), the 22^{nd} (from 4789^{th} year). Between 6840year cycles, month
34 is skipped in 34s only and the interval between skips, not counting this skip of 34, is thirteen 228year cycles. Karl 16(01(20 From: Walter J Ziobro [mailto:[hidden email]]
Dear Karl I thank you for that info I like your second suggestion, that there be 30 leap month shifts, one every 228 years This can be accomplished by skipping a designated leap month every 10 cycles, such that the designated leap month s in the 3334 month groups would be 110, 1221, and
2332 Walter Ziobro Sent from AOL Mobile Mail On Tuesday, September 20, 2016 Karl Palmen <[hidden email]>
wrote: Dear Walter and Calendar People Reply below From: East
Carolina University Calendar discussion List [[hidden email]]
On Behalf Of Walter J Ziobro Dear Karl and Calendar List: KARL REPLIES: Walter’s idea runs into problems after the 33^{rd} month of each 34 or 33
is the leap month. What happens after then? One solution is to jump over the 34^{th} to the 1^{st} to smoothly drop a leap month.
Then 33 of these postponements would be needed in a 6840year cycle, 30 after 11 Metonic cycles and 3 after 3 Metonic cycles so following a subcycle of 120 Metonic cycles of 2280 years. Another idea is to have exactly 30 permitted positions for the leap month in each 34 and 33, then
postponement to the next permitted position would occur exactly once every 12 Metonic cycles of 228 years. Whenever the 30^{th}, 31^{st} or 32^{nd} month in each 33 & 34 were a leap
month, the leap month years would follow the Hebrew 19year cycle. No such cycle would create the smoothest possible distribution of leap months, because some leap
months will occur at least 35 months after the previous, whenever a postponement occurs. I expect it would add a day or two to the solar jitter of about one month. If the leap months were spread as smoothly as possible, we’d have 168month periods of 3433343334
in addition to the 235month periods of 34333433343334 all with fixed leap month. Karl 16(01(19
Original Message Dear Irv & Calendar People I’ve already mentioned an algorithm for cutting calendar cycles where the
leap years are spread as smoothly as possible into symmetrical parts, grouped in larger symmetrical parts. Now I show a similar algorithm where the parts have the leap years as late
as possible. In the first iteration cut after any leap year followed by a common year.
For example: Hebrew 19year cycle: ccLccLcLccLccLccLcL
, Symmetrical 17year leap week cycle:
ccLcccccLcccccLcc
,
33year cycle:
cccLcccLcccLcccLcccLcccLcccLcccLc, Julian 4year cycle:
cccL
Then there should either be two different types of part (joining parts over
end of cycle) or just one part in which case the algorithm has finished.
The part with the greatest proportion of leap years and so shortest if leap years are in minority (shown in
red) is designated as leap year for
the next iteration, even if such parts are in a majority. The examples then become: ccLccLcLccLccLccLcL
, ccLcccccLcccccLcc
,
cccLcccLcccLcccLcccLcccLcccLcccLc The second example has only one double cut and so the algorithm has
finished. If the cycle is started
after the double cut (at 4^{th} year) the leap years will be as late as possible and if the year before the double cut (3^{rd} year) is year Z as defined in earlier notes and if designated year 0, makes K=0.
The third example also has only one double cut and so year Z is the 32^{nd}
year and if started on the 33^{rd} year, the leap years would occurs as late as possible. The first example, which corresponds to the Hebrew 19year cycle, has a
third iteration: ccLccLcLccLccLccLcL It has only one triple cut, which is between the 8^{th} and 9^{th}
year. So the 8^{th} year is year Z and the 9^{th} year is the latest starting year and has subsequent leap years as late as possible.
I may look into Irv’s 353year cycle with 130 leap years and K=269 in a
later note. I’ll say at present that the second iteration cuts it into 11s and 8s. I expect the final cut to occur in the 4^{th} iteration after the 169^{th} year, which I reckoned to be year Z (year with 0 remainder). For leap week calendars of sufficient accuracy, the first iteration cuts
into 6s & 5s, the second iteration into 17s and 11s and the third iteration cuts into 62s and either 45s or 79s, except of course the 62year cycle. The number of iterations that this algorithm runs may be considered to be
a measure of the structural complexity of the cycle. The is also a similar mirror image algorithm, which cuts into parts that
have the leap years as early as possible. Karl 16(01(13

Dear Walter and Calendar People
The original topic concerned cycles whose leap years are spread as smoothly as possible and here I’ll show what happens, if the 2519 leap months of the 6840year
cycle are spread as smoothly as possible. I reckon the 6840year cycle would split into a number of 235month periods of (34333433343334) equal to 19 years and 168month periods of (3433343334) equal to 13 & 7/12 years and a fixed month of each 34
or 33 is a leap month. I also mention similar cycles where the leap months are spread as smoothly as possible. The 6840year cycle has 360*19 years, 360*235  1 = 84599 months and 360*7 – 1 = 25219 leap months. If we count the months from the start of a 6840year cycle, then month M will be a leap month if and only if Remainder of (2519*M + K) divided by 84599 is less than 2519 Over 34 months this remainder goes up by 1047 or down by 845991047.
So if month M has its remainder at least 1047, but less than 25191047=1472, then it is 34 months after the previous leap month and 34 months before the next
leap month and is vice versa. Each period of 235 or 168 months has exactly one such month, which occurs in its first 34. There are 14721047=425 such months and hence 425 periods of 235 or 168 months. If all 425 has 235 months, and so 7 leaps months, we’d have 7*425=2975 leap
months, which is 456 leap months too many and so we have 456/2=228 periods of 168 months and the remaining 425228=197 periods have 235 months. So we have 197 periods (34333433343334) of
235 months = 19 years with 7 leap months & 228 periods (3433343334) of
168 months = 13 & 7/12 years with 5 leap months. Checking: Months: 197*235 + 228*168 = 83599 Leap months: 197*7 + 228*5 = 2519. These would be spread as smoothly as possible over the 6840year cycle.
The number of 235s is roughly equal to the number of 168s, so it may be worth looking at the case, where they are equal. Then we’d have a 403month cycle of
32 & 7/12 years = 33 & 7/12 lunar years, which I recently mentioned in reply to Aristeo.
(3433343334) (34333433343334) of 403 months = 32 & 7/12 years with 12 leap months. Twelve of these make up a 391year cycle equal to 403 lunar months, which is the GrattanGuinness cycle mentioned numerous times before. Other more accurate cycles can be obtained by removing a small number of 235month periods from these twelve 403month cycles, so if leap months are spread
as smoothly as possible producing a few yermlike periods of (168235168235….168). 391 years of 144 leap months: 168235 alternating
372 years of 137 leap months: (168235168235…168) of 23 353 years of 130 leap months: 2*(168235168235…168) of 11 334 years of 123 leap months: 3*(168235168235168235168) of 7 315 years of 116 leap months: 4*(168235158235168) of 5 The 6840year cycle would have 228197=31 of these (168235168235…168) of 13 or 15 spread as smoothly as possible. However the 6840year does not have particular
merit, when the leap months are spread as smoothly as possible and its better suited to earlier suggestions. The following approximations of solar years to lunar years have been alluded to in this note 67 lunar years to 65 solar years in 3433 of 67 months 168 lunar years to 163 solar years in (3433343334) of 168 months 235 lunar years to 228 solar years in (34333433343334) of 235 months 403 lunar years to 391 solar years in (3433343334) (34333433343334) = (168+235) of 403 months Happy Equinox Karl 16(01(21 From: Palmen, Karl (STFC,RAL,ISIS)
Dear Walter and Calendar People Thank you Walter for your reply. I don’t understand what Walter means after “This can be accomplished by skipping”. It needs explaining in more detail. I had suggested the 30 months of each 34 or 33 are permitted leap months as so 4 or 3 months of each 34 or 33 respectively are forbidden months.
These could be (5 15 24 34) in a 34 and (10 19 29) in a 33 forming a Helios cycle over each 34+33, which is interrupted, between 19year cycles when there are
two consecutive 34s. This could be simplified to (4 14 24 34) in a 34 and (9 19 29) in a 33. It could be simplified further to (4 14 24 34) in a 34 and (4 14 24) in a 33.
This further simplification ensures that for any 228year cycle, the leap months occur in the same position (counting from start) in both the 34s and the 33s,
whereas my earlier suggestions would for some 228year cycles, have the leap months occurring 1 month earlier in a 33 than in a 34, assuming they initially occurred at the first month of both. In this simplest suggestion a month would be skipped every
nine 228year cycles within a 6840year cycle starting with the 4^{th} (from 685^{th} year), then the 13^{th} (from 2737^{th} year), the 22^{nd} (from 4789^{th} year). Between 6840year cycles, month
34 is skipped in 34s only and the interval between skips, not counting this skip of 34, is thirteen 228year cycles. Karl 16(01(20 From: Walter J Ziobro [[hidden email]]
Dear Karl I thank you for that info I like your second suggestion, that there be 30 leap month shifts, one every 228 years This can be accomplished by skipping a designated leap month every 10 cycles, such that the designated leap month s in the 3334 month groups would be 110, 1221, and
2332 Walter Ziobro Sent from AOL Mobile Mail On Tuesday, September 20, 2016 Karl Palmen <[hidden email]>
wrote: Dear Walter and Calendar People Reply below From: East
Carolina University Calendar discussion List [[hidden email]]
On Behalf Of Walter J Ziobro Dear Karl and Calendar List: KARL REPLIES: Walter’s idea runs into problems after the 33^{rd} month of each 34 or 33
is the leap month. What happens after then? One solution is to jump over the 34^{th} to the 1^{st} to smoothly drop a leap month.
Then 33 of these postponements would be needed in a 6840year cycle, 30 after 11 Metonic cycles and 3 after 3 Metonic cycles so following a subcycle of 120 Metonic cycles of 2280 years. Another idea is to have exactly 30 permitted positions for the leap month in each 34 and 33, then
postponement to the next permitted position would occur exactly once every 12 Metonic cycles of 228 years. Whenever the 30^{th}, 31^{st} or 32^{nd} month in each 33 & 34 were a leap
month, the leap month years would follow the Hebrew 19year cycle. No such cycle would create the smoothest possible distribution of leap months, because some leap
months will occur at least 35 months after the previous, whenever a postponement occurs. I expect it would add a day or two to the solar jitter of about one month. If the leap months were spread as smoothly as possible, we’d have 168month periods of 3433343334
in addition to the 235month periods of 34333433343334 all with fixed leap month. Karl 16(01(19
Original Message Dear Irv & Calendar People I’ve already mentioned an algorithm for cutting calendar cycles where the
leap years are spread as smoothly as possible into symmetrical parts, grouped in larger symmetrical parts. Now I show a similar algorithm where the parts have the leap years as late
as possible. In the first iteration cut after any leap year followed by a common year.
For example: Hebrew 19year cycle: ccLccLcLccLccLccLcL
, Symmetrical 17year leap week cycle:
ccLcccccLcccccLcc
,
33year cycle:
cccLcccLcccLcccLcccLcccLcccLcccLc, Julian 4year cycle:
cccL
Then there should either be two different types of part (joining parts over
end of cycle) or just one part in which case the algorithm has finished.
The part with the greatest proportion of leap years and so shortest if leap years are in minority (shown in
red) is designated as leap year for
the next iteration, even if such parts are in a majority. The examples then become: ccLccLcLccLccLccLcL
, ccLcccccLcccccLcc
,
cccLcccLcccLcccLcccLcccLcccLcccLc The second example has only one double cut and so the algorithm has
finished. If the cycle is started
after the double cut (at 4^{th} year) the leap years will be as late as possible and if the year before the double cut (3^{rd} year) is year Z as defined in earlier notes and if designated year 0, makes K=0.
The third example also has only one double cut and so year Z is the 32^{nd}
year and if started on the 33^{rd} year, the leap years would occurs as late as possible. The first example, which corresponds to the Hebrew 19year cycle, has a
third iteration: ccLccLcLccLccLccLcL It has only one triple cut, which is between the 8^{th} and 9^{th}
year. So the 8^{th} year is year Z and the 9^{th} year is the latest starting year and has subsequent leap years as late as possible.
I may look into Irv’s 353year cycle with 130 leap years and K=269 in a
later note. I’ll say at present that the second iteration cuts it into 11s and 8s. I expect the final cut to occur in the 4^{th} iteration after the 169^{th} year, which I reckoned to be year Z (year with 0 remainder). For leap week calendars of sufficient accuracy, the first iteration cuts
into 6s & 5s, the second iteration into 17s and 11s and the third iteration cuts into 62s and either 45s or 79s, except of course the 62year cycle. The number of iterations that this algorithm runs may be considered to be
a measure of the structural complexity of the cycle. The is also a similar mirror image algorithm, which cuts into parts that
have the leap years as early as possible. Karl 16(01(13

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