Dear Calendar People
I bring up a topic that I introduced to the list six years ago.
I refer to a cycle of C years or which L years are leap years as a Helios cycle if and only if (1) The cycle is symmetrical in the sense that the first year has the same leap status as the last year, the second year has the same leap status as the penultimate
year etc. (2) The leap years are spread as smoothly as possible. I called such cycles Helios cycles because Helios produced so many examples. The symmetrical 19-year cycle with leap years { 2, 5, 7, 10, 13, 15, 18 } recently mentioned is an example. If L is prime relative to C, then C must be an odd number and year Y is a leap year if and only if ( L*Y + (C-1)/2 ) mod ( C ) < L I considered the issue of when the leap years of one Helios cycle are a subset of the leap years of another Helios cycle of the same length C. For this I believe it is necessary for the number of leap years in one cycle L’ to be a divisor of the number of leap years in the other cycle L. This is not a sufficient condition and fails is L/L’ is an even number. I showed an example in the original note below. I believe it is sufficient for L/L’ to be an odd number, because it is then possible to select every (L/L’)th leap year from the cycle with L leap years in
a manner that preserves the symmetry. This can be done as follows: From the cycle with L leap years, select the ((L/L’+1)/2)th leap year and every (L/L’)th leap year thereafter.
I show this with the example I showed in the original note with C=29, L=6 and L’=2 and so L/L’ =3, which is an odd number. L = 6 { 3, 8, 13, 17, 22, 27 } L’=2 { 8, 22 } One can then see that the leap years of the L’ cycle are the 2^{nd} and 5^{th} leap years of L cycle as obtained by the procedure I gave. Another example is the 19-year cycle L=7 and C=19 already mentioned L=7 { 2, 5, 7, 10, 13, 15, 18 } L’=1 { 10 } For a more interesting example I choose C=41 and L=15 L=15 { 2, 5, 7, 10, 13, 16, 18, 21, 24, 26, 29, 32, 35, 37, 40 } L’=5 { 5, 13, 21, 29, 37 } L’= 3 { 7, 21, 35 } L’ = 1 { 21 } I chose this example, not only because L has two odd divisors, but also C=41 gives the cycle a complicated structure to show my procedure works for a complicated
structure.. C=31 & C=37 each have a simpler structure for L=15. This complexity is reflected in the fact that the example is also the shortest Helios cycle that takes two iterations of my cutting algorithm. This algorithm
cuts the cycle into two types of part, each of which is a Helios cycle and are in turn arranged into a Helios cycle. The second iteration is applied to this second Helios cycle in which the parts of the first iteration are arranged. Each cut is applied exactly
half way between two leap years or parts of the minority type. First iteration cuts are indicated by either ‘|’ & ‘||’ and second iteration cuts are indicated by ‘||’ only. C=41 L=15 { 2 | 5, 7 | 10 || 13 | 16, 18 | 21 | 24, 26 | 29 || 32 | 35, 37 | 40 } The Helios cycle of the second iteration is C=11 L=4 { 2| 5, 7| 10 } which is also the first 11 years of the original cycle. The 293-year cycle of The Symmetry454 Calendar: C=293 L=52 takes only one iteration of my cutting algorithm. The list of leap years at http://individual.utoronto.ca/kalendis/leap/52-293-sym454-leap-years.htm
with the cuts indicated by change in colour. The shorter parts of two leap years occur exactly once every 4 parts (of 62 years) and are all the same colour. One can apply my suggested procedure for L’=4 to get a leap quarter cycle { 37, 110, 184, 257 }. Irv has kindly numbered the leap years to assist in this. Karl Karl 16(01(07 From: Palmen, Karl (STFC,RAL,CICT)
Dear Calendar People One issue that Helios’s recent E-mails have given to me is When is the set of leap years of one Helios cycle a subset of the leap years of another Helios cycle? I’ve have defined a Helios cycle numerous times before on this list and there has been some discussion of it. A call such cycles Helios cycles, because Helios has produced a large number of them, including his most recent a 29-year cycle of six leap years { 3, 8, 13,
17, 22, 27}. A Helios cycle of C years and L leap years has year Y a leap year if and only if ( L*Y + (C-1)/2 ) mod ( C ) < L and C must be an odd number and L be prime relative to C. Such a cycle has its leap years spread as evenly as possible and a certain type of symmetry. Now suppose two Helios cycles have the same number of years C, but a different number of leap years L and L’ and the leap years of the second cycle are all
leap years in the first cycle, then what relationship must L’ have to L? At first I thought L’ would be a divisor or L. Let’s look and see how this works out with C=29 L=1 { 15 } L=2 { 8, 22 } L=3 { 5, 15, 25 } L = 6 { 3, 8, 13, 17, 22, 27 } For L=6 we have L’=2 but not L’=3. Therefore it may be necessary, but not sufficient for L’ to be a divisor of L. We also know that for any odd C, L’=1 for any odd-number L, because the middle year is a leap year if and only if there is an odd-number of leap years. I’ll let other calendar people continue with this issue. Karl 11(06(03 |
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