 Properties of Hebrew Year Periods
Properties of Hebrew Year Periods

by Remy Landau

Properties of Hebrew Year Periods - Part 1

1. Supplementary Conventions

2. Arithmetical 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.

3. Symbolic Conventions

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.

3a. The M+m Assumption

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.

4. The Length of Hebrew Year Periods

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. Evaluating [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}

6. The Lengths of One Hebrew Year

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}

7. The Lengths of One Hebrew Year Including Dehiyyah Lo ADU Rosh

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.

1 Year Lengths Distributions - Under Lo ADU Rosh
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

7.1 The Impact of One Day Postponements

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.

1 Year Lengths Distributions Under A Tuesday Postponement
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 How to Find Starts and Ends of 356 and 382 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 non-allowable 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.

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 non-allowable 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.

Given that some weekday w must be bypassed for the weekday of 1 Tishrei, it is possible to relate to w the weekdays on which the 356 day and 382 day years potentially begin and end. This may be summarized as shown in the following table.

356 DAY YEAR 382 DAY YEAR
DAYSTARTENDSTARTEND
ww + 2dw + 1dw + 1dw - 2d

Let the postponement days be given as w, w + 2d, and w + 4d. Then, according to the relationships tabulated in 7.2.3 above, the starts and ends of the 356 and 382 day years may be tabulated as follows:

356 DAY YEAR 382 DAY YEAR
DAYSTARTENDSTARTEND
ww + 2dw + 1dw + 1dw - 2d
w + 2dw + 4dw + 3dw + 3dw
w + 4dw + 6dw + 5dw + 5dw + 2d

The tabulation shows that in this situation, only w + 6d can start a 356 day year, because w + 2d and w + 4d have been disallowed for year starts.

The table also shows that the only allowable start day for the 382 day year is w + 1d, because days w and w + 2d, the end days of the 382 day year, are not permitted to start any year in this example.

If w = 4, then the postponement days are Wednesday, Friday, and Sunday. Consequently, the only weekday that can start a 356 day year is w + 6d = Tuesday.

And for the same reasons, the only weekday that can then start a 382 day year is w + 1d = Thursday. The day that ends the 382 day year is w - 2d = Monday.

The value of w = 4d was deliberately chosen because it results in the days that are proscribed by Dehiyah Lo ADU Rosh.

If only Wednesday (= w) and Friday (= w + 2d) had been chosen as the proscribed days, then the table shows that the 356 day year could only begin on Sunday, that is, w + 4d = Sunday.

8. Eliminating the 356 Day Hebrew Year
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 2-day 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.

9. Eliminating the 382 Day Hebrew Year

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

L = D" - D' = M + [f+m] + p" - p' (Equation 4.5)

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)
=  2d
Hence, 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 post-leap 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.

9a. All the Lengths of One Hebrew Year

1 YEAR SPANS
12 months =     354d  8h  876p
13 months =     383d 21h  589p
M'+/-DAYSR(D, 7d)OCCURSM"+/-DAYSR(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

10. The 357 Day Year Paradox

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.

10a. An Example Generating The 357 Day Year Paradox

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.

11. The Fixed Calendar's 357 Day Year Impossibility

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' - m
The 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 + 3d
That is why the 357 day year cannot be generated in the fixed Hebrew calendar.

11.1 The Fixed Calendar's 352 Day Year Impossibility

As shown by equation 4.5 above, the length of any period of Hebrew years is given by

L = D" - D' = M + [f+m] + p" - p'

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.

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.

12. Dehiyyah Molad Zakein

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.

The Molad Zakein's Arithmetic Effect

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 (d-m) when z = 18h, and 642p is added to the existing postponement rule limits (z-m) in the Ben Meir calculations
which set z = 18h 642p.

13. Dehiyyah Molad Zakein's Transparency

Dehiyyah Molad Zakein does not affect the statistical distributions of the lengths of any period of Hebrew years.

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.

14. Explaining Dehiyyah Molad Zakein's Transparency

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 (d-z) to the limit (z-m) develops (d-m) 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 => (z-d) + m on a Monday following the end of a 13 month year.

Hence, f + (d-z) => 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 + (d-z) 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'+d-z, T"+d-z) 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., d-z) to the times of the calculated molad, and the limits of the 356 and 382 day elimination rules.

15. L = {M-2d, ..., M+3d} Can NOT Exist for any M + m

No molad period M + m can have both M - 2d and M + 3d as its extreme lengths.

Case 15a. L = M-2d Requires that m < 8h 876p

Equation 4.5 states that L = D" - D' = M + [f+m] + p" - p'.

Hence, in order that a given span of Hebrew years have, as one of its possible lengths, L = M - 2d it is necessary that [f + m] = 0d, p" = 0d, and p' = 2d.

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

According to Equation 4.5, in order that L = M + 3d, it is necessary that
[f+m] = 1d, p" = 2d, and p' = 0d

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 = {M-2d, ..., M+3d} Can NOT Exist for any M + m
In other words, given some molad period M+m, no period of Hebrew years can have both
M - 2d as its minimum and M + 3d as its maximum number of days.

15c. Examples of Property 15

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'+/-DAYSR(D, 7d)OCCURSM"+/-DAYSR(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'+/-DAYSR(D, 7d)OCCURSM"+/-DAYSR(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

16. The Molad Zakein Postponements Produce Neither L = M - 2d Nor L = M + 3d

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.

17. R(M, 7d) When L = M - 2d Or 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}.

17a. An Example of L = M - 2d

2,240 YEAR SPANS
27,705 months = 818,145d  2h  705p
27,706 months = 818,174d 15h  418p
M'+/-DAYSR(D, 7d)OCCURSM"+/-DAYSR(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.

17b. An Example of L = M + 3d

120 YEAR SPANS
1,484 months =  43,823d  9h  692p
1,485 months =  43,852d 22h  405p
M'+/-DAYSR(D, 7d)OCCURSM"+/-DAYSR(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.

17c. NOTE on R(M, 7d) = {0d, 2d, 4d, 6d}

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'+/-DAYSR(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}.

18. Periods for Which L = {M-2d, ...}

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.

18.1 Periods for Which L = {M-2d, ..., M+2d}

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 = M-2d 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.

18a. An Example of L = {M - 2d, ..., M + 2d}

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'+/-DAYSR(D, 7d)OCCURSM"+/-DAYSR(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.

20. The Minimum f and m When L = M + 3d

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.

20a. M + 3d Periods Do Not Start With 354 Day Years

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.

20b. Examples of Periods With L = M + 3d

Example 1 - 19 Hebrew Years

19 YEAR SPANS
235 months =   6,939d 16h  595p
M'+/-DAYSR(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'+/-DAYSR(D, 7d)OCCURSM"+/-DAYSR(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.

Properties of Hebrew Year Periods - Part 2