Si vero mense Aprili Pascha celebramus, computa menses a Septembri usque ad Martium, fiunt VII. His semper adjice II, fiunt IX. Adde epactas lunae anni cujus volueris, ut puta, indictionis IV, XXIII, qui fiunt XXXII, et diem mensis quo Pascha celebramus, id est Aprilis XIX, qui simul fiunt LI; deduc XXX, remanent XXI. Luna XXI est in die resurrectionis Domini.
In the month of April. - If however we celebrate Easter in the month of April, compute the months from September to March, yielding 7. To this always add 2, yielding 9. Add the lunar epacts of the year you want, say 23 for indiction 4, yielding 32, and the day of the month in which we celebrate Easter, that is April 19, which together yield 51; deduct 30, 21 are left over. The age of the moon is 21 on the day of the resurrection of the Lord.
This amounts to the same formula as above for the remaining year numbers: age of the moon on( Julian date(Y, April, D) ) = ( age of the moon on( Julian date(Y, March, 22) ) + 9 + D )mod 30 Thus, the age of the moon is supposed to increase by 1 for each day throughout the 35 day interval from March 21 until April 25 in which these formulae are applicable; this agrees with columns 6 and 8 in the table above. The year number for the example could be 0526.
Si requiras a Septembri usque ad Decembrem, tres semper in his IV mensibus regulares adjicias: in bissexto autem solummodo anno duos regulares suprascriptis mensibus adnumerabis, et pro XXXI die, XXXII annis singulis Decembri mense assumes in fine.
If you need it from September to December, you should always add the correction three in these 4 months: only in a leap year you also shall add the correction two for these months described above, and finally in non-leap years, for day 31 in the month of December you should assume 32.
This is probably meant as a recipe similar to the two above for the age of the moon on( Julian date(Y, January, D) ) = ( (Y mod 19)*11 + 4 + 3 + (1 or 2) + D )mod 30 where the "4" acts as the number of months from September to December, "3" is the the correction in every year, and the "(1 or 2)" comes either from the correction 2 for leap years, or, for non-leap years, it is an interpretation of the effect of assuming 32 days in December.
The interpretation above is consistent with [Argumentum 11] since: Julian date( Y, January, 00 ) ~= Julian date( Y, March, 22 ) - 3 synodic month + (7.6 or 8.6) d (with 8.6 instead of 7.6 for leap year numbers Y).
Argumentum X. De die septimanae sanctae feria paschali.
Si vis cognoscere quotus dies septimanae est, sume dies a Januario usque ad mensem quem volueris, ut puta, ad XXX diem mensis Martii, fiunt LXXXIX. His adjicies semper unum, fiunt XC; et semper adde epactas solis, id est concurrentes septimanae dies cujus volueris anni, ut puta II, indictionis III, fiunt simul XCII. Hos partire per VII, remanet una: ipsa est dominica paschalis festi. Sic quamlibet diem a calendis Januarii usque ad XXXI diem mensis Decembris, quota feria fuerit, invenies computando, ut regularem unum et concurrentes, quae a Januario mense semper incipiunt, pariter assumas.
Argumentum 10. On the day of the holy week of the feast of Easter.
If you want to learn which day of the week it is, add the days since January until the month you want, say until March 30, there are 89. To this always add one, yielding 90; and always add the solar epacts, that is, the concurrents of the seven day week for the year you want, say 2 for the indiction 3, yielding 92 altogether. Divide those by 7, one is left over: this is the Sunday of the feast of Easter. In this way, if you venture to compute which day of the week it is for any day from the first of January until the 31st of the month of December, you should equally assume the correction one and the concurrents which always begin in the month of January.
The example date could be Julian date( 0525, March, 30 ), as can be seen from the table above. The example shows that the number of days from January to( Julian date( Y, January, 01 ) + D d) is meant to be D + 1 rather than D ("Roman inclusive counting"). With the solar epacts of [Argumentum 4], the formula given amounts to day of the week number( Julian date( Y, January, 01) + D d ) = ( D + 1 + 1 + 4 + Y + floor(Y / 4) )mod 7 = ( D + (Y - 1) + floor(Y / 4) )mod 7 which agrees with the correct formula of [Argumentum 12] only if Y is not divisible by 4, and otherwise is one day ahead.
Argumentum XI. De luna citimi paschalis.
Si vis scire quota luna sit in XI calendas Aprilis, sume annos incarnationis Domini nostri Jesu Christi, ut puta, DCLXXV. Hos partire per [ XIX, remanent X; et multiplica decem per ] XI, fiunt CX. Partire tricesima, remanent XX: vicesima luna est in XI calendas Aprilis. Si autem VII, septima; si asse, prima.
Argumentum 11. On the moon closest to Easter. If you want to know which moon it is on March 22, add the years since the incarnation of our Lord Jesus Christ, say 675. Divide those by [ 19, 10 are left over; and multiply ten by ] 11, yielding 110. Divide by 30, 20 are left over: it is the twentieth day of the moon on March 22. And if 7 [is left over], then the seventh, if one, the first.
Only with the suggested correction (and allowing for remainders of zero) this yields age of the moon in year( Y )on March 22 = ((Y mod 19)*11) mod 30 which are the epacts of [Argumentum 3]. Thus, Dionysius Exiguus uses Julian date( Y, March, 22 ) as "sedes epactorum" (seating of the epacts).
Besides these so-called Dionysian epacts, several other epacts have been used in computs for the same or a different Easter date, such as the Alexandrian epacts (8 + (Y mod 19)*11) mod 30, which, according to [Argumentum 13], would give the nominal age of the moon on the day preceding January 01.
Argumentum XII.
Si vis nosse diem calendarum Januarii, per singulos annos, quota sit feria, sume annos incarnationis Domini nostri Jesu Christi, ut puta, annos DCLXXV. Deduc assem, remanent DCLXXIV. Hos per quartam partem partiris, et quartam partem, quam partitus es, adjicies super DCLXXIV, fiunt simul DCCCXLII. Hos partiris per VII, remanent II. Secunda est dies calendarum Januarii. Si V, quinta feria; si asse, dominica; si nihil, sabbatum.
Argumentum 12.
If you want to find out which day of the week it is on the first day of January, for non-leap years, then add the years since the incarnation of our Lord Jesus Christ, say 675 years. Subtract one, 674 are left over. Divide those into the fourth part, and add the fourth part obtained by the division to 674, yielding 842 altogether. Divide those by 7, 2 are left over. It is Monday on the first of January. If 5 [are left over] then [it is] Thursday, if one, then Sunday; if nothing, Saturday.
This amounts to day of the week number of year (Y) on January 01 = ( (Y - 1) + floor( (Y - 1)/4 ) )mod 7 with 0 for Saturday, which is in fact the number for day of the week (Julian date(Y, January, 01)). This is true for leap year numbers Y as well.
Argumentum XIII. De luna calendarum Januarii.
Si vis scire quota luna sit calendis Januarii, scito quotus lunaris cyclus sit, verbi gratia cyclus XV. Tene tibi unum, id est ipsas calendas Januarii, et duces quinquies quinquies decies: faciunt LXXV; quos adjicies super unum, et fiunt LXXVI. Item duces sexies decies quinquies, faciunt XC; quos adjicies super LXXVI, et sic summa numerorum CLXVI; in quibus partiris tricesima, remanent XVI. Sexta decima luna est calendis Januarii, et puncti XVI. Isto modo per XIX cyclos lunares computabis semper, et calendis Januarii, quota sit luna, absque errore reperies.
Argumentum 13. On the age of the moon on the first of January.