Numbers: Their Tales, Types, and Treasures.

Chapter 5: Counting for Poets



One might obtain an idea for a solution when we look at the two groups of meters in table 5.1. The first group has 5; the second, 8 meters. What property distinguishes these two groups? Well, the first group contains all meters that end with a long syllable, while all meters of the second group end with a short syllable. Consider the first group. The parts that precede the long final syllable has a duration of four moras, and the group contains all possible meters with four moras combined with a long syllable. Hence the number of the meters of the first group is just A(4). Similarly, the second group has all possible meters with a duration of five moras, combined with a short final syllable. Hence the number of meters in the second group is A(5). From this we get the formula A(4) + A(5) = A(6) for the number of possible meters with a duration of six moras.

Obviously, we can repeat that reasoning for any n. Any meter with n moras ends with either a long or a short syllable. Thus the set of all meters of length n can be divided into a set of meters that end with a long syllable and a second set of meters that end with a short syllable. The meters ending with a short syllable could start with an arbitrary part of length n – 1 at the beginning; hence there are A(n – 1) such meters. And the meters that end with a long syllable start with an arbitrary part of length n – 2; hence there are A(n – 2) such meters. We conclude that

A(n – 2) + A(n – 1) = A(n)

must hold for any n. Well, at least for n starting with at least 3, so that n – 2 is at least 1. We collect the results of our reasoning:

A(1) = 1, A(2) = 2.
A(n – 2) + A(n – 1) = A(n), for all natural numbers n greater than 2.

Admittedly, this does not directly tell us what A(n) is, but in a way, it solves Pingala's first problem. The formula describes how, starting with the “initial condition” for A(1) and A(2) we can easily compute step-by-step all the numbers A(n):

A(3) = A(1) + A(2) = 1 + 2 = 3,
A(4) = A(2) + A(3) = 2 + 3 = 5,
A(5) = 3 + 5 = 8, A(6) = 5 + 8 = 13, A(7) = 8 + 13 = 21, and so on.

Every further number, A(n) is the sum of the two preceding results. And now we can be sure that we haven't forgotten any of the possible meters of duration 6 in table 5.1, because with the new method we also find A(6) = 13.

With a little patience, we can easily compute the number A(16) to find how many meters have a length of sixteen moras. The old Indian meter “woman with a beautiful body” mentioned earlier is just one of A(16) = 1,597 theoretically possible meters!

A discussion of the numbers A(n) as the solution to Pingala's first problem in verse metrics can be found explicitly in the work of Hemachandra (1089–1172 CE). In the Western world, the sequence of numbers 1, 2, 3, 5, 8, 13, 21, 34…, in which every number greater than 2 is the sum of the two preceding numbers, has been rediscovered quite often. Not knowing that these numbers were already known in old India more than one thousand years earlier, the French mathematician Édouard Lucas (1842–1891) called them Fibonacci numbers after Leonardo da Pisa, more popularly known today as Fibonacci (ca. 1170–ca. 1245). Fibonacci was the most important mathematician in medieval Europe, and he was largely responsible for the popularization of the Indo-Arabic numerals in Europe.