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 ISOlike week rule to the 33years 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 293year 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 
Hi Karl and calendar people, I wrote the attached Perl program (quite a while ago) to calculate exactly such cycles.In order to better see patterns, by default it prints numbers smaller than 62 as a single digit or letters, where 1035 are represented by the letters AZ and 3661 as az. Use the flag "n" to show numbers explicitly.On Tue, Nov 28, 2017 at 2:57 PM, Karl Palmen <[hidden email]> wrote:
 Amos Shapir
leap.pl (1K) Download Attachment 
Dear Amos and Calendar People Thank you Amos for your reply. My Emailer has blocked access to the attachment. Perhaps Amos or some other calendar person could give the output of one or two examples.. My guess for the output of –n 71 400 is something like: 400 231 169 62 62 62 45 62 62 45 17 17 17 11 17 17 17 11 … 6 6 5 6 6 5 6 6 5 6 5 … A more compact possibility is: 400 = 231+169 231 = 62+62+62+45 169 = 62+62+45 62 = 17+17+17+11 45 = 17+17+11 17 = 6+6+5 11 = 6+5 Also the numbers of leap years may also be shown. One important feature about each complexity level is that it contains only
one instance of the minority cycle of the next lower level, hence me italicising
one in the examples. In the case of 71/400 the minority cycle always has the greater mean year and so is shown last. If it were less, I think it would need to be shown first for
consistent ordering. E.g. 334 = 30+19+19+19+…+19 30 = 11+11+8 19 = 11+8 11 = 3+3+3+2 8 = 3+3+2 Karl From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Amos Shapir Hi Karl and calendar people, I wrote the attached Perl program (quite a while ago) to calculate exactly such cycles. Given 2 numbers it prints out the cycles as one line for each complexity level, listing out the lengths of short and long periods. In order to better see patterns, by default it prints numbers smaller than 62 as a single digit or letters, where 1035 are represented by the letters AZ and 3661 as az. Use the flag "n" to show numbers explicitly. The cycle mentioned above would be generated by "perl
leap.pl n 71 400" Have fun,
On Tue, Nov 28, 2017 at 2:57 PM, Karl Palmen <[hidden email]> wrote: 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
ISOlike week rule to the 33years 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 293year 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
Amos Shapir 
Hi Karl and calendar people, Actually the output looks like this:71 400 66566566565665665665656656656656566566565665665665656656656656566566565 HHHBHHHBHHHBHHBHHHBHHHBHHB (62)(62)(62)j(62)(62)j Each line shows the cycle in progressing level of complexity. Note that B, H and j represent 11, 17 and 45, respectively. With the n flags, period lengths on each level are shown explicitly: 71 400 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6 5 17*3 11 17*3 11 17*3 11 17*2 11 17*3 11 17*3 11 17*2 11 62*3 45 62*2 45 Have fun. On Tue, Nov 28, 2017 at 7:23 PM, Karl Palmen <[hidden email]> wrote:
 Amos Shapir

Dear Amos Thank you Amos for your reply and showing example output. I notice that the level
231 169
is not shown. The previous level 62*3 45 62*2 45
cannot be the last before 400, because it has
two of the minority cycle 45. Perhaps, Amos’s program stops at complexity 3 (as 62 & 45 have). I also noticed that when cycles less than 62 years are represented by single characters equal cycles are shown as a sequence rather than multiplied. This would
result in a longer line for 123 334: TIIIIIIIIIIIIIIII 30 19*16 Another thing is if at each level you count the two types of cycles as two types of years then they are spread as smoothly as possible and the complexity of
this cycle plus the complexity of the two types of cycle equal the complexity of the whole cycle. 41/231 & 30/169 have complexity 4 and follow a cycle
1/2
of complexity 1. 11/62 & 8/45 have complexity 3 and follow a cycle
2/7
of complexity 2. 3/17 & 2/11 have complexity 2 and follow a cycle of
7/26
of complexity 3. 1/6 & 1/5 have complexity 1 and follow a cycle of
26/71
of complexity 4. Here I have described the cycles as the fraction giving the proportion of the minority type of ‘years’. If one runs my algorithm for calculating the complexity one gets 71/400
> 45/71 (26/71)
> 19/26 (7/26)
> 5/7 (2/7)
> ½
> 2. The same fractions as in the cycle of cycles occur there. In this example the minority type is always the type with the larger mean year (hence the persistent subtraction from 1 in my algorithm). It would be interesting
to see how Amos’s program behaves in a cycle where this is not so. 97/400 has minority type with smaller mean year for (5 & 4) but larger mean year for (33 & 37). It has complexity 3. 97/400
> 12/97 > 1/12 > 12 Karl 16(16(11 From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Amos Shapir Hi Karl and calendar people, Actually the output looks like this: 71 400 66566566565665665665656656656656566566565665665665656656656656566566565 HHHBHHHBHHHBHHBHHHBHHHBHHB (62)(62)(62)j(62)(62)j Each line shows the cycle in progressing level of complexity. Note that B, H and j represent 11, 17 and 45, respectively. With the n flags, period lengths on each level are shown explicitly: 71 400 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6 5 17*3 11 17*3 11 17*3 11 17*2 11 17*3 11 17*3 11 17*2 11 62*3 45 62*2 45 Have fun. On Tue, Nov 28, 2017 at 7:23 PM, Karl Palmen <[hidden email]> wrote: Dear Amos and Calendar People Thank you Amos for your reply. My Emailer has blocked access to the attachment. Perhaps Amos or some other calendar person could give the output of one or two examples.. My guess for the output of –n 71 400 is something like: 400 231 169 62 62 62 45 62 62 45 17 17 17 11 17 17 17 11 … 6 6 5 6 6 5 6 6 5 6 5 … A more compact possibility is: 400 = 231+169 231 = 62+62+62+45 169 = 62+62+45 62 = 17+17+17+11 45 = 17+17+11 17 = 6+6+5 11 = 6+5 Also the numbers of leap years may also be shown. One important feature about each complexity level is that it contains only
one instance of the minority cycle of the next lower level, hence me italicising
one in the examples. In the case of 71/400 the minority cycle always has the greater mean year and so is shown last. If it were less, I think it would need to be shown first for
consistent ordering. E.g. 334 = 30+19+19+19+…+19 30 = 11+11+8 19 = 11+8 11 = 3+3+3+2 8 = 3+3+2 Karl
Amos Shapir 
Hi Karl and calendar people, The 231 169 level is not shown, because there are no repetitions there; the program does not recognize a pattern and so it stops.On Wed, Nov 29, 2017 at 3:06 PM, Karl Palmen <[hidden email]> wrote:
 Amos Shapir

Dear Amos and Calendar People Thank you Amos for your reply. I wonder how it would cope with 34 89 instead of 71 400. I’d make it (with n flag) 34 89 3*2 2 3*2 2 3 2 3*2 2 3*2 2 3 2 3*2 2 3 2 3*2 2 3*2 2 3 2 3*2 2 3 2 8*2 5 8*2 5 8 5 8*2 5 8 5 21*2 13 21 13 55 34 Karl From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Amos Shapir Hi Karl and calendar people, The 231 169 level is not shown, because there are no repetitions there; the program does not recognize a pattern and so it stops. It is not limited in the number of levels it can show.
On Wed, Nov 29, 2017 at 3:06 PM, Karl Palmen <[hidden email]> wrote: Dear Amos Thank you Amos for your reply and showing example output. I notice that the level
231 169
is not shown. The previous level 62*3 45 62*2 45
cannot be the last before 400, because it has
two of the minority cycle 45. Perhaps, Amos’s program stops at complexity 3 (as 62 & 45 have). 
In reply to this post by Amos Shapir2
Dear Amos, Irv, Victor and Calendar People The cycle for L C, where C is the number of years in the cycle and L the number of leap year has the following leap year rule. Y is a leap year if and only
if (L*Y + K) mod C < L
, where K is a constant dependent on the start year of the cycle. If the leap years occur as late as possible K=0. Also K needs to be 0 for the cycle to divide into the parts listed in the order shown.
I call
(L*Y + K) mod C
the accumulator of year Y. The term accumulator arises from Victor’s 43/450 calendar where he applied the idea to months rather than years.
If K=0, then the last year of the cycle has accumulator 0. Furthermore the last year of each group picked out by Amos’s algorithm and the 231 169 missed out
by it has the smallest accumulator in the group. Here I show some of the groups in L=71 C=400 and the accumulators of their last years in []. 400[0] 231[1]
16 [0] 62[2]
62[4] 62[6] 45[1] 62[3] 62[5] 45[0] 17[7]
17[14] 17[21] 11[2] … 6[26]
6[52] 5[7] 6[33] 6[59] 5[14] 6[40] 6[66] 5[21] 6[47] 5[2] … The groups can be defined such that the last year of each group is its only year Y that satisfies: (L*Y) mod C < N, where N is the number of groups. In the case of L=71, C=400, N is one of
1, 2, 7, 26, 71. I also notice that these numbers appear as the accumulator of the last year of the first group.
Karl 16(16(12 From: East Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Amos Shapir Hi Karl and calendar people, The 231 169 level is not shown, because there are no repetitions there; the program does not recognize a pattern and so it stops. It is not limited in the number of levels it can show.
On Wed, Nov 29, 2017 at 3:06 PM, Karl Palmen <[hidden email]> wrote: Dear Amos Thank you Amos for your reply and showing example output. I notice that the level
231 169
is not shown. The previous level 62*3 45 62*2 45
cannot be the last before 400, because it has
two of the minority cycle 45. Perhaps, Amos’s program stops at complexity 3 (as 62 & 45 have). I also noticed that when cycles less than 62 years are represented by single characters equal cycles are shown as a sequence rather than multiplied. This would
result in a longer line for 123 334:
TIIIIIIIIIIIIIIII 30 19*16 Another thing is if at each level you count the two types of cycles as two types of years then they are spread as smoothly as possible and the complexity of
this cycle plus the complexity of the two types of cycle equal the complexity of the whole cycle. 41/231 & 30/169 have complexity 4 and follow a cycle
1/2
of complexity 1. 11/62 & 8/45 have complexity 3 and follow a cycle
2/7
of complexity 2. 3/17 & 2/11 have complexity 2 and follow a cycle of
7/26
of complexity 3. 1/6 & 1/5 have complexity 1 and follow a cycle of
26/71
of complexity 4. Here I have described the cycles as the fraction giving the proportion of the minority type of ‘years’. If one runs my algorithm for calculating the complexity one gets 71/400
> 45/71 (26/71)
> 19/26 (7/26)
> 5/7 (2/7)
> ½
> 2. The same fractions as in the cycle of cycles occur there. In this example the minority type is always the type with the larger mean year (hence the persistent subtraction from 1 in my algorithm). It would be interesting
to see how Amos’s program behaves in a cycle where this is not so. 97/400 has minority type with smaller mean year for (5 & 4) but larger mean year for (33 & 37). It has complexity 3. 97/400
> 12/97 > 1/12 > 12 Karl 16(16(11 From: East
Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of Amos Shapir Hi Karl and calendar people, Actually the output looks like this: 71 400 66566566565665665665656656656656566566565665665665656656656656566566565 HHHBHHHBHHHBHHBHHHBHHHBHHB (62)(62)(62)j(62)(62)j Each line shows the cycle in progressing level of complexity. Note that B, H and j represent 11, 17 and 45, respectively. With the n flags, period lengths on each level are shown explicitly: 71 400 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6*2 5 6 5 6*2 5 6*2 5 6 5 17*3 11 17*3 11 17*3 11 17*2 11 17*3 11 17*3 11 17*2 11 62*3 45 62*2 45 Have fun. 
In reply to this post by Karl Palmen
Exactly. (Except the last line, 55 34 which has no repetitions)
On Thu, Nov 30, 2017 at 10:55 AM, Karl Palmen <[hidden email]> wrote:
 Amos Shapir

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