If you want to know which moon it is on January 01, knowing which lunar cycle it is, for instance cycle 15. Retain one, which is for the same January 01, and take five fifteen times: yielding 75; to which you always add one, thus yielding 76. Now take six fifteen times, making 90, which you add to 76, thus the sum of the numbers is 166; divide these into the thirtieth [part], 16 are left over. It is the sixteenth moon on January 01, and 16 puncti. In this way you can always compute for the 19 cycles of the moon, and you will obtain without error the age of the moon on January 01.
For year numbers Y, and with the lunar cycle L = 1 + (-3 + Y)mod 19 from [Argumentum 6], this computation yields age of the moon on( Julian date( Y, January, 01 ) ) = ( L*5 + 1 + L*6 )mod 30 = (12 + ((-3 + Y)mod 19)*11 )mod 30 =(for Y mod 19 >= 3:) (9 + (Y mod 19)*11)mod 30 up to to the puncti (to be discussed below).
Unless Y is divisible by 4, this agrees with the formula suggested at the end of [Argumentum 9].
Dum autem veneris ad XVII cycli lunaris, et duxeris quinquies decies septies, super calendas Januarii, qui faciunt LXXXV, si partiris sexagesima, et adjicies ipsum assem, fiunt LXXXVI. Deinde ducis sexies decies septies, fiunt CII. Eos adjicies super LXXXVI, et fiunt CLXXXVIII. [ Adiicies unum, fiunt CLXXXVIIII. ] Partire ibi tricesima, remanent IX. Nona luna est calendis Januarii, et puncti XXVI. Sic et in XVIII et XIX cyclo facies. A primo vero cyclo lunari, usque in sextum decimum, non partiris sexagesimam, ne in errorem incidas.
As soon as you shall come to lunar cycle 17, then take five times seventeen, after January 01, which makes 85, if you divide into the sixtieth [part], and add the resulting one to it, this yields 86. Meanwhile take six times seventeen, yielding 102. Those add to 86, and it yields 188. [ Add one, yielding 189 ]. Divide this by thirty, 9 are left over. It is the ninth moon on January 01, and 26 puncti. In this way you also compute in cycles 18 and 19. From the first lunar cycle until the sixteenth you do not divide by 60 so as not to make an error.
For the Julian year number Y, this computation is said to apply if L = 1 + (-3 + Y)mod 19 is 17, 18, or 19, that is, if L = Y mod 19 + 17. With the addition of 1 as amended above in brackets, it yields age of the moon on( Julian date( Y, January, 01 ) ) = ( L*5 + floor(L/12) + L*6 + 1) mod 30 =(for Y mod 19 < 3:) (9 + (Y mod 19)*11)mod 30 resulting in the same formula as above for the remaining year numbers Y.
Apparently, a separate formula is given for 17 <= L <= 19 because of the term floor(L/12). Of course, floor(L/17) would have worked for all L; this would have required the remainder modulo 85 instead of modulo 60. With the 19 year cycle (as in [Argumentum 5]), a single (and simpler) formula would do.
The separate multiplication by 5 in both computations above is very likely due to a formula of the type fractional age of the moon on( Julian date( Y, January, 01 ) ) = ( A + (Y - B)*30*235/19 )mod 30 = ( A + ((Y - B) mod 19)*(5*(1 + 1/95) + 6) ) mod 30 derived directly from the Metonic value for the synodic month. For integral B and suitable A it gives values for the age of the moon that are integral multiples of 1/95. The number 1/95 is close to a 1/96 = (1 punctus)/(1 d) (see [Argumentum 16]) which would explain the appearance of puncti in the age of the moon.
Unfortunately, the text is not explicit about the computation of the puncti, and the two examples leave many possibilities open, such as: age of the moon on( Julian date( Y, January, 01 ) )in days and puncti = ( 11 + 42/96 + ((Y - 3) mod 19)*(5*(1 + 1/96) + 6) ) mod 30 or = ( 37/96 + ((Y - 2) mod 19)*(5*(1 + 1/96) + 6) ) mod 30 or = ( 19 + 32/96 + ((Y - 1) mod 19)*(5*(1 + 1/96) + 6) ) mod 30 And if we assume that the second example is meant to yield an age of the moon of 8 (rather than 9) plus 16 puncti, then we could have age of the moon on( Julian date( Y, January, 01 ) )in days and puncti = ( 8 + 27/96 + (Y mod 19)*(5*(1 + 1/96) + 6) ) mod 30 In all these formulae, the age of the moon increases by 11 + 5/96 per year except for the "saltus" of 11 + 6/96 once every 19 years.
Thus, this Argumentum incompletely describes a kind of Alexandrian epacts that apparently already had been described more fully elsewhere; I do not know such a source, however.
Argumentum XIV. Quota feria luna XIV incidat cycli decemnovennalis anno primo.
Incipit calculatio quomodo reperiri possit quota feria singularis anni decima quarta luna paschalis, id est primi circuli decemnovennalis.
Argumentum 14. On which day of the week the fourteenth moon falls in the first year of the nineteen year cycle.
The calculation begins whereby one can find out on which day of the week the fourteenth paschal moon falls in a single year, this one being for the first circle of nineteen.
Anno primo, quia non habet epactas lunares, pro eo quod cum noni decimi inferioris anni XVIII, et suis XI epactis, addito etiam ab ¦gyptiis die una, fiunt XXX, id est luna mensis unius integra, et nihil remanet de epactis, et quod in Aprili mense incidit eo anno luna paschalis XIV, tene regulares in eo semper XXXV, subtrahe XXX, id est ipsa luna integra, et remanent V. Quinto die a calendis, hoc est nonis Aprilis, occurrit luna paschalis XIV. Tene suprascriptos V, adde et concurrentes ejusdem anni IV, fiunt IX. Adde et regulares in eodem semper mense Aprili VII, fiunt XVI. Hos partire per VII, id est bis septeni XIV, remanent II. Secunda feria occurrit luna paschalis XIV, et dominicus festi paschalis dies luna XX.
In the first year, which does not have lunar epacts, because to those 18 from the previous nineteenth year, and its 11 epacts, one day is added by the Egyptians, yielding 30, that is one full lunar month, so that nothing remains from the epacts, and so that in this year the 14th paschal moon falls in the month of April, in this year always take the correction 35, subtract 30, that is this full month, and 5 remains. The 14th paschal moon occurs five days from the Kalends, which is April 05. Take the 5 from above, and add the concurrents 4 of this year, yielding 9. And always add to this the correction 7 in the month of April, yielding 16. Divide those by 7, that is, two times seven is 14, 2 are left over. On Monday occurs the 14th paschal moon, and the Sunday of the Easter holiday on the day of the 20th moon.
For Julian year number Y, the rule first given is meant to be day of the 14th paschal moon in year( Y ) = April 35 - ( 16 + (-16 + (Y mod 19)*11 )mod 30 ) d for the special case Y mod 19 = 0. For all integral Y, this is equal to March 21 + ( 15 - (Y mod 19)*11 )mod 30 d which is in fact the first date >= Julian date( Y, March, 21 ) whose age of the moon is 14 according to [Argumentum 11], and using an increase in the age of the moon of 1 mod 30 per day. With the numbering of the days of the week as in [Argumentum 12], the second rule is: ( 35 - 16 - (-16 + (Y mod 19)*11 )mod 30 + day of the week( Julian date( Y, March, 24 ) + 7)mod 7 = ( day of the week( Julian date( Y, March, 24 ) - 3 + ( 15 - (Y mod 19)*11 )mod 30 )mod 7 = day of the week( day of the 14th paschal moon in year( Y ) ) The addition of 7 in this rule is of course unnecessary; it just ensures that X mod 7 is never evaluated with X < 7.
The first sentence describes the "saltus lunae" when the epacts increase by 12 rather than 11 from year (Y - 1) to year Y with Y mod 19 = 0.
Anno secundo.
Item praefati circuli annus secundus, a quo sumunt exordium epactae XI. Incidit in eo anno luna paschalis XIV mense Martii. Tene XXXVI regulares in eo semper, subtrahe semper epactas XI, remanent XXV. Vicesimo quinto die a calendis Martii, quod est VIII calendas Aprilis, occurrit luna paschalis XIV. Tene suprascriptos XXV, adde concurrentes ejusdem anni V, fiunt XXX. Adde semper in fine hujus mensis regulares IV, hos partire per VII, id est septies quaterni XXVIII, remanent VI. Sexta feria occurrit luna XIV paschalis, et dominicus festi paschalis dies luna XVI.
In the second year. Now to the second year of the above mentioned circle, for which the epacts add up to 11 to begin with. In this year, the 14th paschal moon occurs in the month of March. In this [month], always take the correction 36, always subtract the epacts 11, 25 are left over. The 14th paschal moon occurs twenty five days from the beginning of March, that is, on March 25. Take the 25 from above, add the concurrents 5 for this year, yielding 30. Finally always add the correction 4 for this month, divide those by 7, that is four times seven or 28, 6 remain. The 14th paschal moon occurs on Friday, and the Sunday of the feast of Easter is the day of the 16th moon.
For Julian year number Y, the rule given first is day of the 14th paschal moon in year( Y ) = March 36 - ( (Y mod 19)*11 )mod 30 d and if the 14th mooon is in March, then this is again equal to March 21 + ( 15 - (Y mod 19)*11 )mod 30 d because ( (Y mod 19)*11 )mod 30 <= 15 in this case (see [Argumentum 7]). The second rule also amounts to the same as above.
Anno tertio. -
Item mense Aprili saepe dicti circuli primi anno tertio. Tene semper in eo mense imprimis regulares XXXV. Subtrahe epactas ejusdem anni XXII, remanent XIII. Tertio decimo die mensis, id est idibus Aprilis, occurrit luna paschalis XIV. Tene hos XIII, adde concurrentes VI, fiunt XIX. Adde in Aprili semper inferius regulares VII, fiunt XXVI. Hos partire per VII ter septeni, XXI, remanent quinque. Quinta feria erit decima quarta luna paschalis, et dominicus dies paschalis festi luna XVII.
In the third year. In the third year of said first cycle, [the 14th moon occurs] also always in the month of April. For this month, always take first the correction 35. Subtract the epacts 22 of this year, 13 are left over. The 14th paschal moon occurs on the thirteenth day of the month, that is on April 13. Take those 13, add the concurrents 6, yielding 19. Then always add in April the correction 7, yielding 26. Divide those by 7, three times 7 [are] 21, five are left over. On Thursday was the fourteenth paschal moon, and the Sunday of the feast of Easter on the 17th moon.