Dreibens moDulo 7 A New Formula for Primality Testing

towards determInIng whether the set of prIme dreIbens Is InfInIte, we examIne the propertIes of dreIbens when dIvIded by 7. by determInIng the dIvIsIbIlIty of a dreIben by 7, we can rule out some composIte dreIbens In the search for prIme dreIbens. we are concerned wIth the number of dreIbens of length n that leave a remaInder I when dIvIded by 7. by usIng number theory, lInear algebra, and abstract algebra, we arrIve at a formula that tells us how many dreIbens of length n are dIvIsIble by 7. we also fInd a way to determIne the number of dreIbens of length n that leave a remaInder I when dIvIded by 7. further InvestIgatIon from a combInatorIal perspectIve provIdes addItIonal InsIght Into the propertIes of dreIbens when dIvIded by 7. overall, thIs paper helps characterIze dreIbens, opens up more paths of InquIry Into the nature of dreIbens, and rules out some composIte dreIbens from a prIme dreIben search. Dreibens moDulo 7 A New Formula for Primality Testing


IntroductIon to dreIbens and the search for InfInItely many prIme dreIbens
The existence of infinitely many prime numbers is an elementary fact of number theory.However, mathematicians continue to ponder over whether there are infinitely many primes of a certain kind.For instance, it is not yet known whether or not the set of Mersenne primes is finite or infinite.Evidently, questions about primes remain at the forefront of number theory.
Our current research focuses on a type of number that we dubbed dreibens.Dreibens are merely strings of 3s and 7s (e.g.333, 777, 3377373737, etc.), yet these simple strings provide a foundation for a multitude of questions.Our ultimate goal is to prove whether or not infinitely many prime dreibens exist.While that question is beyond the scope of this paper, this paper con tributes to that goal by ruling out some composite dreibens.In fact, this paper concerns itself with dreibens divisible by 7; by understanding dreibens modulo 7, we can rule out some dreibens that are certainly not prime.This paper uses a variety of approaches to learn about dreibens modulo 7, including techniques from number theory, abstract algebra, linear algebra, and combinatorics.With time, we hope dreibens to one day become a very well characterized type of number.
learnIng about dreIbens modulo 7 by ex-plorIng A i 7 (n) We would best approach dreibens by attempting to nd the number of dreibens divisible by some number.For this paper, we interest ourselves in the number of dreibens divisible by 7. Let us define A i j (n) as the number 1 of dreibens of length n that leave a remainder i when divided by j.In this paper, we will concern ourselves with A i 7 (n).How do we find A i 7 (n + 1)?We can find formulas for A i 7 (n + 1) by considering the first n digits of a dreiben of length n + 1, then seeing what the last digit must be such that the dreiben leaves a remainder i when divided by 7. Let us consider a dreiben Dn+1 of length n + 1, which could leave a remainder of 0, 1, 2, 3, 4, 5, or 6 when divided by 7. The rst n digits of Dn+1 form another dreiben Dn that could leave a remainder of 0, 1, 2, 3, 4, 5, or 6 when divided by 7. Hence, we have seven cases, each with seven sub-cases, which we must consider.We will write the rst three cases below.
Notice that in Case 1, Dn+1 covers all possible dreibens of length n + 1 divisible by 7. Likewise, Case 2 covers all possible dreibens of length n + 1 that are 1 (mod 7), Case 3 covers all possible dreibens of length n + 1 that are 2 (mod 7), and so forth.Similarly, each subcase (a) covers all possible dreibens of length n divisible by 7, each subcase (b) covers all possible dreibens of length n that are 1 (mod 7), and so forth.Hence, each case represents A i 7 (n + 1) for some i and each sub-case represents A i 7 (n) for some i.From these cases, we get the following formulas for A i 7 (n + 1) in terms of A i 7 (n): We can rewrite this linear system in terms of matrices.Let The aforementioned linear system is the same as By induction, we find that This linear system allows us to quickly find the number of dreibens of length n that leave a remainder i when divided by 7. We need only calculate the following: Let us investigate A n to see if we nd anything interesting.Evidently:

"We can find formulas for A i 7 (n + 1) by considering the first n digits of a dreiben of length n + 1, then seeing what the last digit must be such that the dreiben leaves a remainder i when divided by 7." 91
Dreibens moDulo 7 If we look at the powers of B and C, we find that Notice how for i = 0 (mod 6), both B i and C i are equal to the identity matrix I7.Both the powers of B and C are cyclic groups of order 6.
Before examining A n , let us define the following: These formulas will be useful for condensing our notation.

Now we can finally begin examining A n :
Now we have The problem with the above equation is that we still have the powers of B in terms of n.However, we can easily solve this problem because the powers of B form a cyclic group.Since this cyclic group has order 6, we must consider six cases.
Evidently, no matter what n is, the first row of matrix A n is always Recall that If we want to find A 0 7 (n + 1), i.e. the first entry of → vn+1, we need only consider the first row of A n : Now we have a formula with which to find the number of dreibens of length n that are divisible by 7: What if we want to find any A i 7 (n), i = 0, 1, 2, 3, 4, 5, 6?We know that these A i 7 (n) are entries of → vn.Recall that To solve for → vn+1, we must first understand A n .Recall that As it turns out, A is a diagonalizable matrix, so A n is also diagonalizable: Interestingly, some entries of D are 6th roots of unity.If we let be the jth ith-root of unity, then

93
Dreibens moDulo 7 Recall that the nth roots of unity form a cyclic of order n, as in Table 1 below: Based on the properties of the 6th roots of unity, we find that Now that we understand D n , we are much closer to characterizing A n .All we need to do is to calculate PD n P -1 for each case of D n .Below, we will calculate the first case, i.e.where n ≡ 0 (mod 6).
If n ≡ 0 (mod 6), Where n ≡ 0 (mod 6), the entries of A n are all either 1/7(2 n -1) or 1/7(2 n + 6).Hence, A n can be much more easily calculated; instead of performing lengthy matrix power operations, one can merely find 1/7(2n -1) and 1/7(2n + 6) to determine A n .Since → vn = A n-1→ v1, one can now easily determine any A i 7 (n) where n ≡ 0 (mod 6).In other words, one can easily determine the number of dreibens of length n ≡ 0 (mod 6) that leave remainder i when divided by 7. By extension, one can determine the number of dreibens of length n ≡ 0 (mod 6) divisible by 7; all of these dreibens are necessarily composite, so they can be ruled out during our search for infinitely many prime dreibens.
We can perform similar calculations for A n where n ≡ 1, 2, 3, 4, 5 (mod 6), thereby allowing us to quickly find A n for any n.Hence, we can easily determine A n for any n, allowing us to rule out many composite dreibens.combInatorIal approach to dreIbens modulo 7 Let us take a combinatorial approach to A i 7 (n).For now, let us restrict ourselves to A 0 7 (n); in other words, we are looking for when a dreiben of length n is divisible by 7.These dreibens are obviously composite, so learning to identify them will help rule out composite dreibens in a prime dreiben search.
Here, we will use the divisibility rule for 7 to aid us.Recall the divisibility rule for 7: Notice that the powers of 10 (mod 7) forms a cyclic group of order 6.
Let us use the above divisibility rule to determine, as an example, 9487310294 (mod 7): Dreibens consist of only 7's and 3's; hence, the divisibility of a dreiben by 7 depends on both the number and placement of 3's amongst its digits.(Neither the number nor placement of 7's in a dreiben affect its divisibility by 7, as we can tell from the divisibility rule for 7.) Let D n j (i) be the number of dreibens of length n with i number of j's in it, e.g.D n

3
(2) is the number of dreibens of length n where 3 occupies exactly 2 digits of each dreiben.Since the divisibility of a dreiben by 7 depends on the number and placement of 3's amongst that dreiben's digits, we would stand to gain from investigating D n 3 (i).
As you may have realized already, Evidently, by examining D n 3 (i), we can approach the problem of determining the number of dreibens of length n divisible by 7. Our approach will involve investigating D n 3 (i) with varying i to see if we can make any generalizations about D n 3 (i).
First, let us examine D n 3 (0).There is only one way to put exactly zero 3's in a dreiben; every digit of that dreiben must be 7. Obviously, if a number's digits are all 7's, then the number is divisible by 7. Since all of the digits of a dreiben with zero 3's is 7, and since there is only one way to make a dreiben of length n with zero 3's, we can easily see that D n 3 (0) = 1.
Next, we can examine D n 3 (1) = 0.There are no dreibens with only one 3, since 3 times any digit place 10 x is never divisible by 7, so D n 3 (1) = 0.So far, we have covered two very simple instances of D n (i) becomes much more complicated.The next question in our investigation of D n 3 (i) involves i = 2.In other words, how many ways are there to put exactly two 3's in a dreiben of length n? Putting exactly two 3's in a dreiben of length n means we must choose two digit places 10 x and 10 y where x ≠ y and 0 ≤ x, y ≤ n -1.Putting our two 3's in these two digit places, we find that the dreiben is congruent to 3(10 x + 10 y ) (mod 7).Clearly, the dreiben is divisible by 7 if and only if 10 x + 10 y is divisible by 7, so we need only consider 10 x + 10 y (mod 7).
What are the possible ways to make 10 x + 10 y ≡ 0 (mod 7)?
The powers of 10 modulo 7 form a cyclic group of order 6; without loss of generality, the possible solutions are For each solution, we choose one digit place each from two sets of digit places.Here, we introduce some notation that will make the later calculations more concise: • α is the set of digit places 10 x ≡ 1 (mod 7), i.e. x ≡ 0 (mod 6).

95
Dreibens moDulo 7 As an example, one possible way to put two 3's into a dreiben is to choose one digit place from α and another digit place from ζ.The number of possible combinations of choosing two digit places, one from α and one from ζ, is ( |α| 1 )( |ζ| 1 ) = |α|•|ζ|.Applying this to the other sets of digit places, we find that As you may have noticed, the cardinality of the six sets above depends on n (mod 6).This can be seen in Table 2 below: Hence, • If n ≡ 0 (mod 6), then Proof.If n ≡ 0 (mod 6), i.e. n = 6m where m is an integer, then If n ≡ 1 (mod 6), i.e. n = 6m + 1 where m is an integer, then If n ≡ 2 (mod 6), i.e. n = 6m + 2 where m is an integer, then If n ≡ 3 (mod 6), i.e. n = 6m + 3 where m is an integer, then conclusIon While we are still far from determining whether the set of prime dreibens is finite or infinite, our exploration of dreibens modulo 7 has cast much light on the problem.Now, we have several methods to determine how many dreibens of length n are divisible by 7, therefore allowing these dreibens to be ruled out during our search for prime dreibens.We have also opened up several opportunities for inquiry.For instance, explicit matrices for A n where n ≡ 1, 2, 3, 4, 5 (mod 6) must still be determined.The relationship between D n 3 (3) and remains a promising mystery, and we have yet to find a generalized formula for D n (a) 10 x ≡ 1 (mod 7), 10 y ≡ 6 (mod 7) (b) 10 x ≡ 2 (mod 7), 10 y ≡ 5 (mod 7) (c) 10 x ≡ 3 (mod 7), 10 y ≡ 4 (mod 7) Another way to write the possible solutions is (a) x ≡ 0 (mod 6), y ≡ 3 (mod 6) (b) x ≡ 2 (mod 6), y ≡ 5 (mod 6) (c) x ≡ 1 (mod 6), y ≡ 4 (mod 6)