In applying any sequence of names or numbers to another there are only three possibilities concerning the names or numbers which can stand at the head of the resulting sequence:
1. When the sums of the units in each of the two sequences contain no common factor, each one of the units in turn will stand at the head of the resulting sequence.
2. When the sum of the units in one of the two sequences is a multiple of the sum of the units in the other, only the first unit can stand at the head of the resulting sequence.
3. When the sums of the units in the two sequences contain a common factor (except in those cases which fall under (2), that is, in which one is a multiple of the other) only certain units can stand at the head of the sequence.
Now, since our two numbers (the 20 names in Table [I] and the 365 days of the year) contain a common factor, and since neither is a multiple of the other, it is clear that only the last of the three contingencies just mentioned concerns us here; and we may therefore dismiss the first two from further consideration.
The Maya year, then, could begin only with certain of the days in Table [I], and the next task is to find out which of these twenty names invariably stood at the beginnings of the years.
When there is a sequence of 20 names in endless repetition, it is evident that the 361st will be the same as the 1st, since 360 = 20 × 18. Therefore the 362d will be the same as the 2d, the 363d as the 3d, the 364th as the 4th, and the 365 as the 5th. But the 365th, or 5th, name is the name of the last day of the year, consequently the 1st day of the following year (the 366th from the beginning) will have the 6th name in the sequence. Following out this same idea, it appears that the 361st day of the second year will have the same name as that with which it began, that is, the 6th name in the sequence, the 362d day the 7th name, the 363d the 8th, the 364th the 9th, and the 365th, or last day of the second year, the 10th name. Therefore the 1st day of the third year (the 731st from the beginning) will have the 11th name in the sequence. Similarly it could be shown
that the third year, beginning with the 11th name, would necessarily end with the 15th name; and the fourth year, beginning with the 16th name (the 1096th from the beginning) would necessarily end with the 20th, or last name, in the sequence. It results, therefore, from the foregoing progression that the fifth year will have to begin with the 1st name (the 1461st from the beginning), or the same name with which the first year also began.
This is capable of mathematical proof, since the 1st day of the fifth year has the 1461st name from the beginning of the sequence, for 1461 = 4×365+1 = 73×20+1. The 1 in the second term of this equation indicates that the beginning day of the fifth year has been reached; and the 1 in the third term indicates that the name-part of this day is the 1st name in the sequence of twenty. In other words, every fifth year began with a day, the name part of which was the same, and consequently only four of the names in Table [I] could stand at the beginnings of the Maya years.
The four names which successively occupied this, the most important position of the year, were: Ik, Manik, Eb, and Caban (see Table [V], in which these four names are shown in their relation to the sequence of twenty). Beginning with any one of these, Ik for example, the next in order, Manik, is 5 days distant, the next, Eb, another five days, the next, Caban, another 5 days, and the next, Ik, the name with which the Table started, another 5 days.