1. Supplementary Conventions
2. Arithmetical Conventions
3. Symbolic Conventions
3a. The M+m Assumption
4. The Length of Hebrew Year Periods
5. Evaluating [f+m] + p"  p'
6. The Lengths of One Hebrew Year
7. The Lengths of One Hebrew Year Including
Dehiyyah Lo ADU Rosh
7.1 The Impact of One Day Postponements
7.2 How to Find Starts and Ends of 356 and 382 Day Years
7.2.1  Starts and Ends of 356 Day Years
7.2.2  Starts and Ends of 382 Day Years
7.2.3  Summary of Starts and Ends of 356 and 382 Day Years
7.2.4  Starts and Ends of 356 and 382 Day Years Given w, w+2, w+4
8. Eliminating the 356 Day Hebrew Year
9. Eliminating the 382 Day Hebrew Year
9a. All The Lengths of One Hebrew Year
10. The 357 Day Year Paradox
10a. An Example Generating The 357 Day Year Paradox
11. The Fixed Calendar's 357 Day Year Impossibility
11.1 The Fixed Calendar's 352 Day Year Impossibility
12. Dehiyyah Molad Zakein
13. Dehiyyah Molad Zakein's Transparency
14. Explaining Dehiyyah Molad Zakein's Transparency
15. L = {M2d, ..., M+3d} Can NOT Exist for any M + m
Case 15a. L = M2d Requires that m < 8h 876p
Case 15b. L = M+3d Requires that m > 15h 204p
15c. Examples of Property 15
16. The Molad Zakein Postponements Produce
Neither L = M  2d Nor L = M + 3d
17. R(M,7d) When Either L = M  2d Or L = M + 3d
17a. An Example of M  2d
17b. An Example of M + 3d
17c. NOTE on R(M, 7d) = {0d, 2d, 3d, 6d}
18. Periods for Which L = {M2d, ...}
18.1 Periods for Which L = {M2d, ..., M+2d}
18a. An Example of L = {M  2d, ..., M + 2d}
20. The Minimum f and m When L = M + 3d
20a. M + 3d Periods Do Not Start With 354 Day Years
20b. Examples of Periods With L = M + 3d
247 Hebrew Year Periods
10 Lengths Hebrew Year Periods
34 Day Variance Hebrew Year Periods
These conventions are added to the ones listed in Hebrew Calendar Science and Myths Conventions.
2.1 The square bracketed expression [x] will mean the INTEGER portion of x.
For example, [3.1416] = 3, [.25] = 0, 2.718  [2.718] = .718, etc...
2.2  Definition of R(a, K)
Given any integers a, A, and K
a + A * K = R(a, K) whenever 0 <= a + A * K <= K – 1.
Algebraically,
R(a, K) = a  [a / K] * K whenever 0 <= a  [a / K] * K <= K  1.
2.3 The braced expression {a, b, ...,z} will be used to indicate the set of all of the possible values that may be selected for a particular variable or expression.
For example, R(D, 7d) = {0, 2, 3, 5} implies that the
week day D may be either
Saturday (0), Monday (2), Tuesday (3), or Thursday (5).
2.4 Common elements within braces may be reduced as follows {a, ..., a, ...} = {a, ..., ...}
For example, {1, 2, 3, 1, 5, 3} = {1, 2, 3, 5}
2.5 The expression {a, b, ...,z} + {x , y, ... , } evaluates to
{a+x, a+y, ... b+x, b+y, ... ... z+x, z+y, ... }
Hence, the expression {a,b} + {c,d} + {e,f} evaluates to
{a+c, a+d, b+c, b+d} + {e,f}
which results in
{a+c+e, a+c+f, a+d+e, a+d+f, b+c+e, b+c+f, b+d+e, b+d+f}
2.6
Let X = { x(1), x(2), x(3), .... } Let Y = { y(1), y(2), y(3), .... } Let Z = { z(1), z(2), z(3), .... } Then for some x in X, some y in Y, and some z in Z, the equation X + Y = Z will have as its solution all of the pairs (x , y) for which x + y = z.
As an example, suppose that Y  X = 3, and that Y = {6, 7, 8, 9}, then the pairs (x, y) that could be solutions to the the equation are (3, 6), (4, 7), (5, 8) and (6, 9).
Other considerations would have to be made in order to determine whether or not any of {3, 4, 5, 6} are to be found in the set X.
I am leaving alone any more general definitions or discussions of the ideas
presented by
Arithmetical Conventions 3, 4, 5, and 6,
since I believe that the use of these conventions in the following text
is reasonably easy to understand.
T' represents the molad of Tishrei for year H'.
T" represents the molad of Tishrei for year H".
T"  T' = the period of time between two known moladot of Tishrei.
Let T"  T' = M + m
where
M = the integer portion of the time period,
ie, [M+m]
m = the fractional portion of that period.
Hence, m = M+m  [M+m]
Thus for 12 Hebrew months, M = 354d and m = 8h 876p for 13 M = 383d and m = 21h 589p for 19 Hebrew years M = 6939d and m = 16h 595p
For convenience, the following symbols represent these values
d = 24h 0p; d' = 23h 1079p; d" = 24h 1p c = 8h 876p; c' = 8h 875p; c" = 8h 877p g = 15h 204p; g' = 15h 203p; g" = 15h 205p b = 21h 589p; b' = 21h 588p; b" = 21h 590p f = any fractional part of a day such that 0 <= f <= d' p' = the number of days that Tishrei 1 is postponed at the start of the year p" = the number of days that Tishrei 1 is postponed at the end of the year T' = the molad of Tishrei of some year H' T" = the molad of Tishrei of some subsequent year H" D' = the day of 1 Tishrei H' D" = the day of 1 Tishrei H"
It is to be noted that
i) d  g = 8h 876p and d  b = 2h 491p
ii) p' and p can be {0, 1, 2} days.
Computer analysis demonstrates that, over the full Hebrew calendar cycle of 689,472 years, each of the possible 181,440 values for the Tishrei Moladot, when reduced to the form R(M, 7d) + m, is repeated exactly 3 or exactly 4 times.
The reduced form of the Tishrei Moladot for the 11th, 13th, and 15th years of the mahzor katan (19 year cycle) known as GUChADZaT are the only ones that are repeated exactly 3 times over the full Hebrew calendar cycle of 689,472 years.
It is therefore reasonable to assume, that for any given M + m, there exists at least one period of Hebrew years whose lunar length in reduced form is R(M, 7d) + m.
As an example, given the value 2d 5h 204p,
we will assume that there exists at least one period of
Hebrew years, no matter how long, whose lunar length M + m, when
reduced to the form
R(M,7d) + m, will be 2d 5h 204p.
The time of the molad T' for year H' is given by 4.1 T' = [T'] + f.If some postponement rule applies to [T'], then the day for 1 Tishrei H' is
4.2 D' = [T'] + p'
The molad of Tishrei for year H" is given by T" = T' + M + m = [T'] + f + M + m (substituting equation 4.1 for T') Hence 4.3 [T"] = [T'] + M + [f+m] If some postponement rule applies to [T"], then the day for 1 Tishrei H" is 4.4 D" = [T'] + M + [f+m] + p" The length L between years H" and H' is given by subtracting equation 4.2 from equation 4.4, yielding
4.5 L = D"  D' = M + [f+m] + p"  p'
5.1 [f+m] = {0d, 1d} , since (0d <= f < d) and (0d <= m < d). 5.2 When p' and p" = {0d, 1d} then [f+m] + p"  p' = {0d, 1d} + {0d, 1d}  {0d, 1d} = {0d, 1d, 1d, 2d}  {0d, 1d} = {0d, 1d, 1d, 0d, 1d, 0d, 2d, 1d} = {1d, 0d, 1d, 2d} 5.3 When p' and p" = {0d, 1d, 2d} then [f+m] + p"  p' = {0d, 1d} + {0d, 1d, 2d}  {0d, 1d, 2d} = {0d, 1d, 2d, 1d, 2d, 3d}  {0d, 1d, 2d} = {0d, 1d, 2d, 3d}  {0d, 1d, 2d} = {0d, 1d, 2d, 1d, 0d, 1d, 2d, 1d, 0d, 3d, 2d, 1d} = {2d, 1d, 0d, 1d, 2d, 3d}
Absent the Dehiyyot, (ie, the postponement rules) the length L of a single Hebrew year is
L = M + [f+m] = {354d, 383d} + {0d, 1d} (Since M = {354d, 383d} and 5.1 above) = {354d, 355d, 383d, 384d}
Lo ADU Rosh requires that the first day of Tishrei
be postponed to the next day whenever
R([T],7d) = {1d, 4d, 6d}.
Consequently, both D' and D" = {0d, 2d, 3d, 5d}, and both p'and p" = {0d, 1d}.
Under this rule, the length of one Hebrew year can become
L = M + [f+m] + p"  p' (By equation 4.5) = {354d, 383d} + {0d, 1d} + {0d, 1d}  {0d, 1d} = {354d, 383d} + {1d, 0d, 1d, 2d} (by equation 5.2 above) = {353d, 354d, 355d, 356d, 382d, 383d, 384d, 385d}
The actual distribution of these lengths, over the complete Hebrew calendar cycle of 689,472 years, is shown in the following table.
YEAR LENGTH IN DAYS  

DAY  353  354  355  356  382  383  384  385  TOTALS 
Mon  39369  0  85047  0  0  40000  0  32576  196992 
Tue  0  39369  0  22839  0  0  36288  0  98496 
Thu  0  101577  22839  0  3712  36288  0  32576  196992 
Sat  39369  0  85047  0  0  40000  0  32576  196992 
TOTALS  78738  140946  192933  22839  3712  116288  36288  97728  689472 
The 356 day and 382 day years are produced whenever a single day is removed from the days of the week which are permitted for 1 Tishrei. It does not matter which day is chosen. This effect is predicted by Equations 4.5 and 5.1.
Now, let w be the day not allowed for 1 Tishrei.
In the following example, Tuesday is removed from the allowable weekdays, while the remaining 6 days are allowed. The actual distribution of the single years lengths, over the complete Hebrew calendar cycle of 689,472 years, is shown in the following table.
YEAR LENGTH IN DAYS  

DAY  353  354  355  356  382  383  384  385  TOTALS 
Sun  0  39369  22839  0  0  3712  32576  0  98496 
Mon  0  39369  22839  0  0  3712  32576  0  98496 
Wed  39369  62208  22839  0  3712  36288  0  32576  196992 
Thu  0  39369  0  22839  0  0  36288  0  98496 
Fri  0  0  62208  0  0  3712  32576  0  98496 
Sat  0  39369  22839  0  0  3712  32576  0  98496 
TOTALS  39369  219684  153564  22839  3712  51136  166592  32576  689472 
7.2.1  Starts and Ends of 356 Day Years
Let w = the weekday disallowed for 1 Tishrei. Let M + m = 354d 8h 876p and L = 356d.Then L = D"  D' = 354d + [f + m] + p"  p' = 356d. (Equation 4.5) Hence, [f + m] + p"  p' = 356d  354d = 2d. Consequently, it is necessary that p" = 1d.
Leading to D" = w + p" = w + 1d. Hence, w + 1d  D' = 356d D' = w  355d R(D', 7d) = R(w  355d, 7d) = R(w  5d, 7d) = R(w + 2d, 7d)
Therefore, given the single nonallowable weekday w, the 356 day year will begin on D' = w + 2d and end on D" = w + 1d.
In the above example, with Tuesday removed, the 356 day year is seen to begin on Tuesday + 2d = Thursday. It will end on Tuesday + 1d = Wednesday since R(Thursday + 356d, 7d) = Wednesday.
7.2.2  Starts and Ends of 382 Day Years
Let w = the weekday disallowed for 1 Tishrei. Let M + m = 383d 21h 589p and L = 382d.Then L = D"  D' = 383d + [f + m] + p"  p' = 382d. (Equation 4.5) Hence, [f + m] + p"  p' = 382d  383d = 1d. Consequently, it is necessary that p' = 1d.
Leading to D' = w + p' = w + 1d. Hence, D"  w  1d = 382d D" = w + 383d R(D", 7d) = R(w + 383d, 7d) = R(w + 5d, 7d) = R(w  2d, 7d)
Therefore, given the single nonallowable weekday w, the 382 day year will begin on D' = w + 1d and end on D" = w  2d.
In the above example, with Tuesday removed, the 382 day year is seen to begin on Tuesday + 1d = Wednesday. It will end on Tuesday  2d = Sunday since R(Wednesday + 382d, 7d) = Sunday.
7.2.3  Summary of Starts and Ends of 356 and 382 Day Years
356 DAY YEAR  382 DAY YEAR  

DAY  START  END  START  END 
w  w + 2d  w + 1d  w + 1d  w  2d 
7.2.4  Starts and Ends of 356 and 382 Day Years Given w, w+2, w+4
356 DAY YEAR  382 DAY YEAR  

DAY  START  END  START  END 
w  w + 2d  w + 1d  w + 1d  w  2d 
w + 2d  w + 4d  w + 3d  w + 3d  w 
w + 4d  w + 6d  w + 5d  w + 5d  w + 2d 
Since R(D"  D', 7d) = R(356d, 7d) = 6d, R(D', 7d) = R(D"  6d, 7d) = R({0d, 2d, 3d, 5d}  6d, 7d) = {1d, 3d, 4d, 6d} = {3d} due to Dehiyyah Lo ADU Rosh
Of the 4 values possible for R(D', 7d), only 3d (Tuesday) is allowed.
Therefore, the 356 day year begins on Tuesday, and ends on R(3d+356d, 7d) = 2d (Monday).
Let the molad period of 12 Hebrew months be expressed as 354d + m.
Since the length of any given period of Hebrew years is given by
L = D"  D' = M + [f+m] + p"  p' (Equation 4.5)
and there is yet no 2day postponement, the 356 day year occurs
whenever
f + m > d' , p" = 1, and p' = 0.
So that [f+m] = 1d, it is necessary that f+m > d' (or f > d'  m).
When the molad of Tishrei for year H' is on
Tuesday past d'm,
postponing 1 Tishrei H' to Thursday forces
p' = 2d, thus making the length of the 12 month Hebrew year H'
354 days because
L = D"  D' = 354d + [f+m] + p"  p' (By equation 4.5) = 354d + 1d + 1d  2d = 354d
Therefore, whenever the molad of Tishrei for a 12 month year occurs past 15h 203p (ie, d'  m) on a Tuesday, postponing the start of Tishrei to Thursday eliminates the 356 day year.
To see how the value 15h 203p varies as a result of the
Molad Zakein's timing, please refer below to
Dehiyyah Molad Zakein
Let the molad period of 13 Hebrew months be expressed as 383d + m, where m = 21h 589p.
Since the length of any given period of Hebrew years is given by
the 382 day year occurs whenever f + m < d , p" = 0, and p' = 1.
Since Dehiyyah Molad Zakein has not yet been introduced, and the year is leap, p' cannot be 2d.
Now, R(D"  D', 7d) = R(382d, 7d) = 4d, R(D', 7d) = R(D"  4d, 7d) = R({0d, 2d, 3d, 5d}  4d, 7d) = {3d, 5d, 6d, 1d} = {3d, 5d} due to Dehiyyah Lo ADU Rosh
Since p' = 1, and Lo ADU Rosh does not postpone
1 Tishrei from
Monday (2d) to Tuesday (3d),
but does postpone 1 Tishrei from
Wednesday (4d) to Thursday (5d), D' = {5d}.
Consequently, R(D", 7d) = R(D' + 382d, 7d) = R(5d + 4d, 7d) = 2dHence, when
the molad of Tishrei H' is on Wednesday (4d), and M = 383d, and f+m < d, implying f < d  m then p' = 1, year H" begins on Monday (2d), making p" = 0 (due to Lo ADU Rosh), and the length of year H = M + [f+m] + p"  p' (Equation 4.5) = 383d + 0d + 0d  1d = 382d
Since, the earliest time f, for any molad of Tishrei year H, is 0h 0p, the smallest value of f+m for year H+1, following a leap year, is 21h 589p.
That time on Monday (2d) would cause p" to remain zero,
and year H to be 382 days unless
1 Tishrei H+1
is postponed to Tuesday (3d) whenever a postleap year molad
of Tishrei is on Monday (2d) past 21h 588p.
In so doing, p" becomes 1 thereby making equation 4.5
D"  D' = 383d + [f + 21h 588p] + 1d  1d = 383d + 0 (since it is given that f + 21h 588p < 1d)leading to 383 days for the length of leap year H.
To see how the value 21h 588p varies as a result of the
Molad Zakein's timing, please refer below to
Dehiyyah Molad Zakein
1 YEAR SPANS  

12 months = 354d 8h 876p 
13 months = 383d 21h 589p 

M'+/  DAYS  R(D, 7d)  OCCURS  M"+/  DAYS  R(D, 7d)  OCCURS  
2  0d  0  0  2  0d  0  0  
1  353d  3  69,222 
1  0d  0  0  
0  354d  4  167,497 
0  383d  5  106,677 

1  355d  5  198,737 
1  384d  6  36,288 

2  0d  0  0  2  385d  0  111,051 

3  0d  0  0  3  0d  0  0  
The maximum variance is 32 days 
If the rule to eliminate 356 day years is applied when M = 354d, [f+m] = 1d, p" = 2d and p' = 0d then according to equation 4.5 L = 354d + 1d + 2d  0d = 357d . The 357 day year does not occur in the fixed Hebrew calendar. The paradox can be explained as follows. To eliminate the 356 day years, a postponement to Thursday (5d) is made whenever the fractional part of the molad of Tishrei on a Tuesday (3d) is greater than d'm. The fractional part of the molad is given by f + m  [f+m]. Since f + m  [f+m] > d' m and [f+m] = d, (by hypothesis) f + m  d > d' m Hence f + m > d + (d' m) and m > d + (d' m)  f The inequality shows that the minimum value for m will be when f is at its maximum value which is d' (by definition). Hence, the minimum value of m required to develop a 357 day year is given by m > d + (d' m)  d' m > d  m m > d / 2 In the fixed Hebrew calendar, the actual value of m for the 12 month year is 8h 876p, which is less than d/2. That is why the 357 day year cannot be generated in the fixed Hebrew calendar.
ASSUMING that the molad period had been 29d 13h 1p then the 12 month year would have been 354d 12h 12p.
The 356 day could then have been eliminated by postponing Rosh Hashannah to Thursday (5d) whenever the molad of Tishrei of a 12 month year fell on Tuesday (3d) past 11h 1067p.
This postponement would then have caused the year to be 354d.
However, suppose that in this scenario, the molad of Tishrei of a 12 month year occurred on Thursday (5d) at 23h 1056p.
Then 5d 23h 1056p + 354d 12h 12p = 360d 11h 1068p
Now R(360d 11h 1068p, 7d) = 3d 11h 1068p
Since this time of the molad of Tishrei is Tuesday (3d) past 11h 1067p, it is necessary to postpone Rosh Hashannah to Thursday (5d), thereby adding 2 days to the year.
Consequently, the final sum of days is 362d.
Since the year began at 5d, the length of the year is
362d  5d = 357 days.
Therefore, if the fractional part of the length of 12 Hebrew months had been greater than 12 hours, the creation of a 357 day year would have been an unexpected problem arising from the elimination of the 356 day year.
In the fixed Hebrew calendar, the actual value of m for the 12 month year is 8h 876p, which is less than d/2.
When m < d / 2 2m < d m < d  m When m < d  m and [f+m] = d f + m < d + d'  m (since the maximum f = d') f + m  [f+m] < d'  mThe fractional part of the sum f+m is less than the value called for in the 356 day postonement rule. Consequently, it is not possible to have p" = 2d when [f+m] = d and m < d/2.
Therefore, [f+m] + p" < 3d when m < d/2
Since L = 354d + [f+m] + p" L < 354d + 3dThat is why the 357 day year cannot be generated in the fixed Hebrew calendar.
As shown by equation 4.5 above, the length of any period of Hebrew years is given by
If the rule to eliminate 356 day years is applied when
M = 354d, [f+m] = 0d, p" = 0d and p' = 2d then according to equation 4.5 L = 354d + 0d + 0d  2d = 352d .However, the 352 day year never occurred in the fixed Hebrew calendar, and is not possible, for the following reasons.
To eliminate the 356 day years, a postponement to Thursday (5d) is made whenever the fractional part of the molad of Tishrei on a Tuesday (3d) is greater than d'  m. (See 8. Eliminating the 356 Day Hebrew Year).
Thus, such a postponement can only take place when f > d'  m.
Now, in order that [f+m] = 0d, as assumed above, it is necessary that f + m < d, in which case, f < d  m.
Since it is not possible simultaneously to have both f < d  m and f > d'  m, therefore, it is not possible to generate 352 day years.
Dehiyyah Molad Zakein limits the maximum calculated time of the molad to 23h 422p of the first day of any new month. This purpose is fully explained in Understanding the Molad Zakein Rule.
The Molad Zakein rule requires that 1 Tishrei
be postponed to the next allowable day whenever
f => 18h on an allowable day for Rosh Hashannah.
An unsuccesful attempt to increase that limit to 18h 642p was made by Aaron Ben Meir in 920g (4680H). For more information on this attempt, please refer to The Ben Meir Years.
Let z be the time that triggers the Molad Zakein postponement for some [T] = {0d, 2d, 3d, 5d}.
For reasons similar to those shown in
Eliminating the 356 Day Hebrew Year it is necessary that
f + m < z so as not to call on the postponement
that eliminates the 356 day year.
Hence, f < z  m
Let zlimit = z  m
Let dlimit = d  m (as derived above)
Then, zlimit = dlimit + z  d
The same relationship is developed between zlimit and dlimit when dealing with the postponement limit that eliminates the 382 day year.
That's why 6 hours is subtracted from the postponement rule limits
(dm) when z = 18h, and 642p is added to the
existing postponement rule limits (zm) in the Ben Meir
calculations
which set z = 18h 642p.
Dehiyyah Molad Zakein does not affect the statistical distributions of the lengths of any period of Hebrew years.
This fact is observed for lengths of one Hebrew year in The Keviyyot.
Let T + f be the time of a molad of Tishrei under the
Molad Zakein rule.
Then T + f + (d  z) will lead to the same day for
Tishrei 1 in the absence of the Molad Zakein rule.
This relationship explains why the Molad Zakein rule introduces no additional lengths, whatever, to any period of Hebrew years.
Let R([T], 7d) = {0d, 2d, 3d, 5d} and f => z,
then under the Molad Zakein rule,
R(D, 7d) = {2d, 3d, 5d, 0d}
If there is no Molad Zakein rule, and f => z, then f + (d  z) => d.
Hence, when R([T], 7d) = {0d, 2d, 3d, 5d},
then R([T], 7d) + [f + (d  z)] = [T] + d = {1d, 3d, 4d, 6d}
Consequently, R(D, 7d) = {2d, 3d, 5d, 0d} due to LO ADU Rosh
When R([T], 7d) = {1d, 4d, 6d}, under the Molad Zakein rule, R(D, 7d) = {2d, 5d, 0d} due to LO ADU Rosh
If R([T], 7d) = {1d, 4d, 6d}, absent the Molad Zakein rule,
then R([T], 7d) + [f + (d  z)] = [T] + d = {2d, 5d, 0d}.
Let R([T], 7d) = {0d, 2d, 3d, 5d} and f < z.
Then under the Molad Zakein rule,
R(D, 7d) = {0d, 2d, 3d, 5d}.
If R([T], 7d) = {0d, 2d, 3d, 5d} and f < z,
then, absent the Molad Zakein rule,
R([T], 7d) + [f + (d  z)] = [T] + 0d = {0d, 2d, 3d, 5d}
Under the Molad Zakein rule, the postponement which eliminates the 356 day year is triggered whenever f => z  m on a Tuesday at the start of a 12 month year.
For the start of a 12 month year,
Let R([T], 7d) = {3d} and f => z  m.
Then, under the Molad Zakein rule, R([T], 7d) = {5d}
Adding (dz) to the limit (zm) develops (dm) which is the postponement time threshold limit in the absence of the Molad Zakein rule (as shown above).
Also, under the Molad Zakein rule, the postponement which eliminates the 382 day year is triggered whenever f => (zd) + m on a Monday following the end of a 13 month year.
Hence, f + (dz) => m which times cause the postponement to take place in the absence of the Molad Zakein rule.
Consequently, T + f under the Molad Zakein rule produces the same result as T + f + (dz) in the absence of the Molad Zakein rule.
Hence, every pair (T', T") under the Molad Zakein rule has a one to one correspondence to every pair (T'+dz, T"+dz) absent the Molad Zakein rule.
Since the length of any given time period is defined by the pair (T', T"), there exists a one to one correspondence between all of the lengths of all of the time periods calculated either with or without the Molad Zakein rule.
Therefore, the Molad Zakein rule introduces no new time lengths between the Tishrei moladot and we can ignore the Molad Zakein rule in relation to calculating the lengths of any period of Hebrew years.
The above explains why the Molad Zakein rule is transparent to the statistical distribution of the lengths of any period of Hebrew years.
Karl Friedrich Gauss, the 18th century mathematical genius, bypassed the Molad Zakein rule in his well known Pesach formula by adding 6 hours (i.e., dz) to the times of the calculated molad, and the limits of the 356 and 382 day elimination rules.
That formula, given any Hebrew year, automatically calculates the Julian date of Pesach for that year. The formula is shown in The Gauss Pesach Formula.
No molad period M + m can have both M  2d and M + 3d as its extreme lengths.
Case 15a. L = M2d Requires that m < 8h 876p
Equation 4.5 states that L = D"  D' = M + [f+m] + p"  p'.
In the absence of Dehiyyah Molad Zakein, p' = 2d only when the 356d elimination rule is applied to a Tuesday (3d) because f > g'. (See 8. Eliminating the 356 Day Hebrew Year).
Now g' < f (so as to cause a 356 day elimination postponement) Hence g' + m < f + m < d (since [f+m] = 0) and g' + m < d' (since max f+m = d' and g'+m < f+m) Therefore, m < d'  g' or, m < 23h 1079p  (15h 203p) = 8h 876p
Case 15b. L = M+3d Requires that m > 15h 204p
In the absence of Dehiyyah Molad Zakein, p" = 2d only when the 356d elimination rule is applied to a Tuesday (3d) because f+m  [f+m] > g'. (See 8. Eliminating the 356 Day Hebrew Year).
Since, f+m  [f+m] > g' and [f+m] => d f+m > d + g' m > d  f + g' > d  d' + g' = g (substituting maximum f = d')Hence, the minimum value of m > g = 15h 204p (by convention for g)
Therefore, it is necessary that m > 15h 204p whenever L = M + 3d.
Cases 15a and 15b above require that m < 8h 876p in order that L = M  2d exist and that m > 15h 204p in order that L = M + 3d exist Therefore, L = {M2d, ..., M+3d} Can NOT Exist for any M + mIn other words, given some molad period M+m, no period of Hebrew years can have both
The 4 year span is the smallest period of Hebrew years containing
the 5 lengths ranging from
M  1d to M + 3d on the left hand side of the table.
Because the left hand side of the table contains the length M + 3d it cannot have any length corresopnding to M  2d.
4 YEAR SPANS  

49 months = 1,446d 23h 1057p 
50 months = 1,476d 12h 770p 

M'+/  DAYS  R(D, 7d)  OCCURS  M"+/  DAYS  R(D, 7d)  OCCURS  
2  0d  0  0  2  0d  0  0  
1  1,445d  3  78  1  1,475d  5  52,504 

0  1,446d  4  74,676 
0  1,476d  6  18,230 

1  1,447d  5  249,668 
1  1,477d  0  255,858 

2  1,448d  6  32,762 
2  0d  0  0  
3  1,449d  0  5,696 
3  0d  0  0  
The maximum variance is 32 days 
The 5 year span is the smallest period of Hebrew years containing
the 5 lengths ranging from
M  2d to M + 2d on the left hand side of the table.
Because the left hand side of the table contains the length M  2d it cannot have any length corresopnding to M + 3d.
5 YEAR SPANS  

61 months = 1,801d 8h 853p 
62 months = 1,830d 21h 566p 

M'+/  DAYS  R(D, 7d)  OCCURS  M"+/  DAYS  R(D, 7d)  OCCURS  
2  1,799d  0  14  2  0d  0  0  
1  1,800d  1  9,856 
1  1,829d  2  30,531 

0  1,801d  2  67,945 
0  1,830d  3  155,318 

1  1,802d  3  25,353 
1  1,831d  4  225,267 

2  1,803d  4  5,696 
2  1,832d  5  169,492 

3  0d  0  0  3  0d  0  0  
The maximum variance is 33 days 
Let z = the Molad Zakein's postponement threshold. Let f => z when R(D', 7d) = {0d, 5d}. Then p' = {1d, 2d}. Since f => z, f + m => z Hence, p" = {1d, 2d}. Since L = M + [f+m] + p"  p' (equation 4.5) L = M + {0d, 1d} + {1d, 2d}  {1d, 2d} = M + {1d, 2d, 2d, 3d}  {1d, 2d} = M + {0d, 1d, 1d, 0d, 1d, 0d, 2d, 1d} = M + {1d, 0d, 1d, 2d} Now, let f < z when R(D', 7d) = {0d, 5d}. Then p' = 0d. And f + m < z + m < z + d causing no Molad Zakein postponement when [f+m] = d. Hence, p" = {0d, 1d, 2d} when [f+m] = 0d and p" = {0d, 1d} when [f+m] = d Consequently, when f < z either L = M + 0d + {0d, 1d, 2d} = M + {0d, 1d, 2d} or L = M + 1d + {0d, 1d} = M + {1d, 2d}
Therefore, the Molad Zakein postponements produce neither L = M  2d nor L = M + 3d.
When L = D"  D' = M  2d p' = 2d and R(D', 7d) = 5d (due to the 356 day year elimination rule) Hence, R(D"  5d, 7d) = R(M  2d, 7d) R(M, 7d) = R((D"  3d, 7d) = {0d, 2d, 3d, 5d}  3d (since R(D", 7d) = {0d, 2d, 3d, 5d}) = {4d, 6d, 0d, 2d} Similarly, when L = D"  D' = M + 3d, p" = 2d and R(D", 7d) = 5d (due to the 356 day year elimination rule) R(5d  D', 7d) = R(M + 3d, 7d) R(M, 7d) = R(2d  D', 7d) = 2d  {0d, 2d, 3d, 5d} (since R(D', 7d) = {0d, 2d, 3d, 5d}) = {2d, 0d, 6d, 4d}Therefore, when L = M  2d or L = M + 3d, and the 356 day year is eliminated,
R(M, 7d) = {0d, 2d, 4d, 6d}.
2,240 YEAR SPANS  

27,705 months = 818,145d 2h 705p 
27,706 months = 818,174d 15h 418p 

M'+/  DAYS  R(D, 7d)  OCCURS  M"+/  DAYS  R(D, 7d)  OCCURS  
2  818,143d  4  14,902 
2  0d  0  0  
1  818,144d  5  188,276 
1  0d  0  0  
0  818,145d  6  52,303 
0  818,174d  0  103,015 

1  818,146d  0  252,551 
1  818,175d  1  19,055 

2  0d  0  0  2  818,176d  2  59,156 

3  0d  0  0  3  818,177d  3  214  
The maximum variance is 34 days 
2240 Hebrew years can span a molad period of M = 818,145d 2h 705p.
R(M, 7d) = R(818,145d, 7d) = 6d
and the smallest possible L = M  2d = 818,143d.
120 YEAR SPANS  

1,484 months = 43,823d 9h 692p 
1,485 months = 43,852d 22h 405p 

M'+/  DAYS  R(D, 7d)  OCCURS  M"+/  DAYS  R(D, 7d)  OCCURS  
2  0d  0  0  2  0d  0  0  
1  43,822d  2  114,296 
1  43,851d  3  1,404 

0  43,823d  3  188,409 
0  43,852d  4  21,423 

1  43,824d  4  167,629 
1  43,853d  5  96,784 

2  43,825d  5  73,986 
2  43,854d  6  19,332 

3  0d  0  0  3  43,855d  0  6,209 

The maximum variance is 33 days 
120 Hebrew years can span a molad period of M = 43,852d 22h 405p.
R(M, 7d) = R(43852d, 7d) = 4d
and the largest possible L = M + 3d = 43,855d.
The fact that R(M, 7d) = {0d, 2d, 4d, 6d}
does not imply the existence of either L = M  2d
or L = M + 3d.
247 YEAR SPANS  

3,055 months = 90,215d 23h 175p 

M'+/  DAYS  R(D, 7d)  OCCURS 
2  0d  0  0 
1  90,214d  5  10,317 
0  90,215d  6  3,439 
1  90,216d  0  675,716 
2  0d  0  0 
3  0d  0  0 
The maximum variance is 2 days 
247 Hebrew years have the molad period of 90,215d 23h 175p.
Although R(M, 7d) = R(90215, 7d) = 6d, the only lengths possible
for the period of
247 years are
L = {M  1d, M, M + 1d} = {90214d, 90215d, 90216d}.
See also 20. The Minimum f and m When L = M + 3d.
For a given number of Hebrew years of lunar period M + m, the length M  2d can exist only when 0h <= m < 8h 876p.
27. Year Spans That Exclude L" = M"  2d discusses the surprising conditions which will deny the length M  2d, even though M + m satisfies Properties 17 and 18 above.
For a given number of Hebrew years of lunar period M + m, the lengths M  2d and M + 2d can exist when R(M, 7d) = {0d, 2d, 4d} and 0h <= m < 8h 876p.
However, these conditions do not garantee that the given span of Hebrew years will have both the lengths L = M2d and L = M+2d.
It seems that when R(M, 7d) = 4d the inequality should be 2h 490p <= m < 8h 876p in order to have the period of years also include the length L = M + 2d.
No period of Hebrew years can include both the lengths
L = M  2d and L = M + 2d when
R(M, 7d) = 6d.
However, when the conditions R(M, 7d) = 6d AND 2h 491p < m < 8h 876p are found for a given span of Hebrew years, then the difference that can exist between the shortest and longest length for that span is the maximum possible 34 days.
This remarkable property is proved in Properties of Hebrew Year Periods  Part 2
40 Hebrew years form a period for which
L = {M  2d, ..., M + 2d}.
One of the 2 molad periods possible for the period of
40 years is M + m = 14,588d 2h 782p.
40 YEAR SPANS  

494 months = 14,588d 2h 782p 
495 months = 14,617d 15h 495p 

M'+/  DAYS  R(D, 7d)  OCCURS  M"+/  DAYS  R(D, 7d)  OCCURS  
2  14,586d  5  6,574 
2  0d  0  0  
1  0d  0  0  1  14,616d  0  87,101 

0  14,588d  0  168,691 
0  14,617d  1  53,449 

1  14,589d  1  291  1  14,618d  2  298,958 

2  14,590d  2  5,884 
2  14,619d  3  68,524 

3  0d  0  0  3  0d  0  0  
The maximum variance is 33 days 
In this example, it can be seen that R(M, 7d) = R(14588d, 7d) = 0d.
Property 17. R(M, 7d) When Either L = M  2d Or L = M + 3d shows that
R(M, 7d) = {0d, 2d, 4d, 6d} whenever L = M + 3d.
It is also required that m > 15h 204p whenever L = M + 3d.
Similarly, absent the Molad Zakein rule, it can be shown that the minimum value for f = 15h 205p whenever L = M + 3d. Otherwise, the minimum value for f = 15h 205p + z  d.
It is interesting to note that the above also explains why periods of Hebrew years that have the length L = M + 3d can never start with 354 day years.
354 day years begin on Tuesday (3d). M + 3d periods require that f > 15h 204p which time on such Tuesdays would trigger the 356 day year elimination postponement thus causing p' = 2d instead of the required p' = 0d. And that, in turn, would make M + 3d impossible according to equation 4.5.
Example 1  19 Hebrew Years
19 YEAR SPANS  

235 months = 6,939d 16h 595p 

M'+/  DAYS  R(D, 7d)  OCCURS 
2  0d  0  0 
1  6,938d  1  11,263 
0  6,939d  2  311,544 
1  6,940d  3  250,123 
2  6,941d  4  113,011 
3  6,942d  5  3,531 
The maximum variance is 4 days 
19 Hebrew years have the
molad period M + m = 6939d 16h 595p,
thus leading to R(M, 7d) = 2d, and m = 16h 595p > 15h 204p.
The longest periods are L = M + 3d = 6942 days, occurring 3,531 times over the full Hebrew calendar cycle of 689,472 years.
Example 2  120 Hebrew Years
120 YEAR SPANS  

1,484 months = 43,823d 9h 692p 
1,485 months = 43,852d 22h 405p 

M'+/  DAYS  R(D, 7d)  OCCURS  M"+/  DAYS  R(D, 7d)  OCCURS  
2  0d  0  0  2  0d  0  0  
1  43,822d  2  114,296 
1  43,851d  3  1,404 

0  43,823d  3  188,409 
0  43,852d  4  21,423 

1  43,824d  4  167,629 
1  43,853d  5  96,784 

2  43,825d  5  73,986 
2  43,854d  6  19,332 

3  0d  0  0  3  43,855d  0  6,209 

The maximum variance is 33 days 
One of the 2 molad periods for
120 Hebrew years is M + m = 43,852d 22h 405p,
thus leading to R(M, 7d) = 4d, and m = 22h 405p > 15h 204p.
The longest periods are L = M + 3d = 43,855 days, occurring 6,209 times over the full Hebrew calendar cycle of 689,472 years.
None of these periods begins with a 354 day year.
First Paged 16 Apr 2001 Next Revised 20 Apr 2003