Pearl necklaces for Maheśvara

Śrīpati’s pearl necklace for Maheśvara
The brāhmaṇa Śrīpati of the Kāśyapa clan was a soothsayer from Rohiṇīkhaṇḍa, which is in the modern Buldhana district of Maharashtra state. Somewhere between 1030 to 1050 CE he composed several works on mathematics, astronomy and divination, one of which is the Gaṇita-tilaka on basic arithmetic and algebra which has come down to us through incomplete manuscripts. In that he posses the following problem (The solution is provided by his Jaina commentator Siṃhatilaka Sūri in 1275 CE):

viśva-kha-sapta-bhujaṅga-navārkāḥ śaila-turaṅga-samāhata-dehāḥ |
syāt sphuṭa-tāraka-vartula-muktā-bhūṣaṇam atra maheśvara-kaṇṭhe ||
The bodies of all (13), the space (0), the seven (7), the snakes (8), the nine (9) and the suns (12) struck together (means multiplied in mathematical terminology) with the mountains (7) and the horses (7) may now be the clear, sparkling and globular pearls forming an ornament for the neck of Maheśvara.

The purpose of the problem is two-fold: (1) to make the student familiar with using the Hindu numerical code (the bhūta-saṃkhya) and (2) make the student do some elementary large number multiplication. Thus, the problem is actually a simple multiplication 12987013 \times 77 whose answer is 1000000001. Interestingly, Śrīpati offers a clue for the answer in the problem itself: he says the multiplication of the 2 numbers results in a necklace of clear, sparkling, spherical pearls for Maheśvara. This is clearly an allusion to the palindromic structure of the product, with the 0s forming the pearls, since they were written historically as circles and the two flanking 1s form the bindings of the necklace. Some numerical savants are known to exhibit a synesthesia with respect numbers — perhaps such was indeed the situation with Śrīpati for a palindromic number such as this could simultaneously produce in ones mind the vivid image of a pearl necklace.

Moreover, this is not just any pearl necklace but one for Rudra’s neck. It is in this regard we believe he encoded more into that number. As we can see from the above, one of the factors of 1000000001 is 11, which is the characteristic number of the Rudra-s. At the same time, the Rudra-s are also said to be 1000s upon 1000s (Yajurveda: sahasrāṇi sahasraśo ye rudrā adhi bhūmyām |). Thus, this product captures both those aspects. Further, the numbers that yield the product are described in bhūta-saṃkhya (itself eminently amenable synesthetic experience of numbers) as including viśvā (= all); kha (= space); 7, which symbolizes the heavenly realms or vyāhṛti-s, the 8 directional earth-bearing snakes, the 9, which symbolizes the planets, the 12 Āditya-s on one hand and the 7 continental mountain ranges and 7 solar horses on the other. Thus, the two numbers are described by the entities of entire universe pervaded by the 1000s of Rudra-s with their 11-fold essence and their product is seen yielding a necklace for Maheśvara. In this regard, the use of viśva for 13 is curious. In the bhūta-saṃkhya system, viśva represents the viśvedeva-s (all gods). In the gaṇapāṭha database from which Pāṇini constructed his grammar we find viśvadeva as entry 13 in the manojñādi-gaṇa (GP 177.13), thus lending viśva to encode 13.

Maheśvara’s necklace sequence and its factors
Taking the cue from Śrīpati, we can define a general integer sequence f[n] of Maheśvara’s necklaces thusly: f[n]=10^n+1, where n is an integer and f[n]^m, where m=1,2,3,4, is also palindromic. This implies that n=1, 2, 3... Thus, the first few terms of our sequence are:
11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001, 10000000001…

We can next ask what are the factors of each f[n]. The first few f[n] are factorized and tabulated below:

Table 1

From the above one may notice a few obvious things. The first 2 terms of the sequence are primes, 11 and 101, but all others are composites. It is also obvious that 3 can never be a factor of f[n] because the sum of the digits of f[n] will never be 3. We also observe that many of the f[n] have a tendency to have a mix of small factors with a very large one. Śrīpati’s original example f[9] is one such: f[9]=7\times 11\times 13\times 19\times 52579. We can then ask questions such as: 1) which primes will divide a given f[n]; 2) For which n will a given prime p be a factor of f[n]. 3) Knowing these, we can ask questions, such as, at what further n will we get a f[n] that will be again divisible by 52579, i.e. they will be further Maheśvara’s necklaces of the type specified by Śrīpati.

We notice right away that 11 divides every other term, i.e. whenever n is odd. We also notice that 101 is a factor of f[2] and every 4th term thereafter. Thus, we can formally write that 11 divides every f[n] when n=2k+1, where k=0,1,2,3.... Similarly, 101 divides every f[n] when n=4k+2. Thus, it also becomes obvious that 11 and 101 will never be co-factors of the same f[n]. With closer observation we can see that every prime p that divides a subset of f[n] does so at some n=m\cdot k+\tfrac{m}{2}, where m=2,4,6,8...: the reason for writing it this way will become clear below. The first few p that divide f[n] for some n are tabulated in the order of how often they do so along with the form n takes when f[n] is divisible by that p:

Table 2

At first site the order in which the primes which divide f[n] appear beyond 11 and 101 is puzzling — they wildly differ in magnitude and form. However, a closer examination reveals a striking pattern behind this: a prime p appears in the above list as per the multiplicative order of 10 modulo p: consider 10^j \mod p where j=1, 2, 3...; when for the first time 10^j \mod p = 1, that j is m the multiplicative order of 10 \mod p. Carl Gauss had famously shown in his Disquisitiones Arithmeticae that m is the length of the repeat pattern of the decimal expansion of \tfrac{1}{p}. Thus, it is also clear that p divides 10^j-1 for the first time when j=m. For example, for 7 we get m=6 because 7 divides 10^j-1 for the first time when j=6 to give \tfrac{999999}{7}=142857. Likewise, the repeat pattern in the decimal expansion of \tfrac{1}{7}=0.\overline{142857} which is of length 6. From the above we can easily see why a p will divide f[n] first time when n=\tfrac{m}{2}. Thus, the sequence of p that divide f[n] for the first time will be arranged as per the multiplicative order of 10 \mod p:

10^m-1= 10^{(m/2)^2}-1=(10^{m/2}-1)(10^{m/2}+1)

Now p divides 10^j-1 for the first time when j=m. Hence, it will not divide 10^{m/2}-1. However, because p divides 10^m-1, it therefore divides 10^{m/2}+1

Now what if p divides some factor of 10^{m/2}+1 which takes the form 10^j+1; j=1, 2, 3...? We can see from polynomial factorization that a polynomial of the form x^j+1 can frequently have two factors of the form x^j+1; j=1, 2, 3..., namely x+1 or x^2+1. For example, x^3+1=(x+1)(x^2-x+1) and x^6+1=(x^2+1)(x^4-x^2+1). Now, in our case x+1 \equiv 11 and x^2+1 \equiv 101. Those are primes and the first two terms of f[n]; hence, they will not be divided by any other p. Now, less frequently, other numbers of the form 10^j+1 are divisors of another such number for a larger j. For example, 10001 is a divisor of 1000000000001. So, let us assume for a moment that p divides some 10^l+1 which is a factor of 10^{m/2}+1, then l<\tfrac{m}{2}. If p does divides 10^l+1, then it also divides 10^{2l}-1. But 2l-1<m; hence, p cannot divide it because it will only divide a number of of the form 10^j-1 when j=m, i.e. the multiplicative order of 10 \mod p. Thus, p cannot divide any other 10^l+1 where l<\tfrac{m}{2}.

The Maheśvara’s necklaces f[n] are the sequence 10^j+1; therefore they would be divided by a given p for the first time when n=\tfrac{m}{2}, which is half the multiplicative order of 10 \mod p \;_{...\blacksquare}

By the procedure we followed above we can see that, after n=\tfrac{m}{2}, p would divide every f[n] where n=m\cdot k+\tfrac{m}{2}; k=1, 2, 3... A corollary to this is only p with even m can be factors of f[n] for only then \tfrac{m}{2} would be an integer. Hence, those primes with odd m such as 3, m=1; 37, m=3 etc will never be factors of any f[n].

Armed with the above, we can also tell which will be the next f[n] that will have 52579 as a factor as the original example of Śrīpati. For 52579, m=18 \; \therefore p | f[n] \iff n=18k+9; \; k=0,1,2... Hence, next term would be:

We can also see that some p will always come together as factors of f[n] because they have the same m. Thus, 7 and 13 with m=6 or 19 and 52579 with m=18 will always co-occur. Further, if a certain n satisfies the relationship n=m \cdot k +\tfrac{m}{2} for a certain p and p also divides that n then p will occur again as a factor of f[n]. For example, consider n=6 \times 3+3=21. Now, m=6 here; hence f[21] will be divisible by both 7 and 13 as they have m=6. However, 7 divides 21. Hence, 7 will occur again as the factor of f[21]. Thus, we have: f[21]= \underline{7} \times \underline{7} \times 11 \times 13 \times 127 \times 2689 \times 909091 \times 459691. Likewise, n=6 \times 6 +3=39 will correspond to a f[n] divisible by both 7 and 13. However, as 39 is divisible by 13, we will have 13 occur again as a factor of f[39]. Thus, f[39]=7 \times 11 \times \underline{13} \times \underline{13} \times 157 \times 859 \times 6397 \times 216451 \times 1058313049 \times 388847808493. All other f[n] would be square-free.

The families of Maheśvara’s necklaces

Figure 1

We can represent any given f[n] as a clique of its factors. For example, Figure 1 shows the dodecagonal clique formed by the factors of f[45], which the most composite f[n] for n=1..50. We then merge all cliques sharing common nodes for f[n], n=1..50. The edges and the nodes are then scaled as per their frequency of occurrence across all 50 cliques. The result is a factor graph for f[n] which is shown in Figure 2.

Figure 2 (click on figure to magnify)

We can see from Figure 2 that there are totally 6 families of f[n] in this range. These families can be described according to their founder member which is then the divisor of the remaining f[n] of that family. The founder member of each family can be described as the f[n] of the form 10^{2^l}+1, l=0,1,2..., where 2^l corresponds to a particular \tfrac{m}{2}:

● When l=0, \tfrac{m}{2}=1, we get the 11 family. 11, as we saw above, divides every f[n] corresponding to n=2k + 1. Thus, every f[n] corresponding to an odd \tfrac{m}{2} is drawn into this family, there making it the largest of them.
● When l=1, \tfrac{m}{2}=2, we get the 101 family. 101 draws all f[n] corresponding to n=4k+2. Thus it becomes the largest of the even \tfrac{m}{2} families.
● When l=2, \tfrac{m}{2}=4, we get the 1001 family. 1001 being composite is centered on its two factors 73 and 137 and corresponds to the terms where n=8k+4.
● When l=3, \tfrac{m}{2}=8, we get the 100000001 family. This number being composite is centered on its factors 17 and 5882353 and corresponds to the terms where n=16k+8.
● When l=4, \tfrac{m}{2}=16, we get the 10000000000000001 family centered on its factors 353, 449, 641, 1409 and 69857. This encompasses the terms corresponding to n=32k+16.
● When l=5, \tfrac{m}{2}=32, we get the 100000000000000000000000000000001 family centered on its factors 976193, 19841, 6187457 and 834427406578561. This includes the terms corresponding to n=64k+32.

Thus, we find that the even \tfrac{m}{2} terms are split up among the various families that appear as per the powers of 2.

The largest factor of Maheśvara’s necklace
Given the above information, we can cut down the time in which we factorize Maheśvara’s necklaces and gather the set of factors for the first 300 terms of f[n]. We can then ask which is the largest prime p_m[n] which divides the corresponding f[n]. Figure 3 shows the plot of \log_{10}p_m[n] against \log_{10}(f[n]).

Figure 3

We see that the general increase of \log_{10}(p_m) appears to be linear with \log_{10}(f[n]). It is bounded between lines y=a_ux, y=a_lx, where a_u=1 and a_l \approx 0.1831. The upper bounding slope a_u is easy to understand: as we observed above, some f[n] tend to have factors widely differing in magnitude; thus the large one is closer in magnitude to f[n]. Trivially, first two terms are primes. There after we get f[n] that are minimally composite. These tend to be of a particular form, e.g.:
f[19]=11 \times 909090909090909091
f[31]= 11 \times 909090909090909090909090909091
f[53]= 11 \times 9090909090909090909090909090909090909090909090909091
Thus, these have factors that approach the upper bounding line.

The median value of ratio of p_m[n] to f[n] is approximately 0.5006. This indicates an even distribution with half the number of p_m[n] being greater than the \sqrt{f[n]} and the other half being lesser than \sqrt{f[n]}.

We understand the lower bound is less clearly. Is there a way to derive it from theory alone? One can see that for n corresponding to multiples of 15 there is an increased propensity to be close to the lower bound. This is in part expected from the factorization of polynomial of the form x^n+1 where n is a multiple of 15. For example, we can see that:

x^{13}+1=(x + 1) (x^{12} - x^{11} + x^{10} - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) \\[5pt] x^{14}+1=(x^2 + 1) (x^{12} - x^{10} + x^8 - x^6 + x^4 - x^2 + 1) \\ [5pt] x^{15}+1=(x + 1) (x^2 - x + 1) (x^4 - x^3 + x^2 - x + 1) (x^8 + x^7 - x^5 - x^4 - x^3 + x + 1) \\ [5pt] x^{16}+1=x^{16} + 1 \\ [5pt] x^{17}+1=(x + 1) (x^{16} - x^{15} + x^{14} - x^{13} + x^{12} - x^{11} + x^{10} - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)

As one can see from the above example the polynomial where n=15 tends to have more factors than other polynomial adjacent to it. Thus, these tend to be among highly composite f[n]; hence, they are more likely to have smaller p_m[n] than their neighbors.

Finally, let us look at some actual historical pearl ornaments of Maheśvara. The below were likely ātmaliṅga-s of a medieval South Indian ruler or a royal śaiva officiant.

Figure 4

In this example we see the first term of Maheśvara necklace sequence. 11 is of course the characteristic number of the Rudra-s. If we wish to add the bottom two pearls which belong to a different register we get 13 which is the next most frequent factor of f[n].

Figure 5

In this example we get two sets of pearl ornamentation one with 17 and another with 7. Both of these are factors of f[n], with 7 being the next most frequent and 17 the founder of a distinct family. Of course there are other numbers with other symbolisms in these ornaments.

For background also see:
Fermat’s little theorem and the periods of the reciprocals of primes
Visualizing the Hindu divisibility test

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