RSC pertains to squarefree root of base n

icarus
Dozens Demigod
icarus
Dozens Demigod
Joined: 12:29 PM - Apr 11, 2006

5:02 PM - Aug 02, 2018 #1

This morning I had a realization that perhaps is not earth shaking and might seem esoteric, but pertains to useful aspects of number bases.

Let's begin with some definitions.

Let the infinite tensor C include all numbers coprime to base n, meaning gcd(k, n) = 1 for all k in C. It follows that 1 and all primes q that do not divide n are in C, and all products of any primes q to any positive integer multiplicity also are in C. Therefore it is plain to see that the tensor C is unbounded and infinite. Example: the tensor C pertaining to uncial is {1, 5, 7, b, 11, 15, 17, 1b, 21, 25, 27, 2b, ...}. It is easy to see a pattern. C contains all numbers k such that k (mod n) ≡ t in T, where T is the (necessarily finite) set of totatives t of n (i.e., the reduced residue system of n).

Let the infinite tensor R include all "regular" numbers r such that r | n^e with e ≥ 0. It is easy to see that the divisor d | n is a special case of r where 0 ≤ e ≤ 1. Therefore R contains 1 and the other divisors d, as well as any number r that divides some power of n. Since we can always increment e to obtain a larger power n^e, and since those numbers have some divisors r (at least n^e itself) that do not divide n^(e − 1), by induction we show R to be infinite. Example: the tensor R that pertains to uncial is {1, 2, 3, 4, 6, 8, 9, 10, 14, 16, 20, 23, 28, 30, ...}; this contains 1 and the other divisors of the dozen (2, 3, 4, 6, 10) and numbers that only divide the dozen to some integer power greater than 1 (8, 9, 14, 16, ...).

The intersection of R and C is {1}, since 1 | n and gcd(1 , n) = 1; this is a unique property of the number 1. Outside of this term, R and C are distinct but not coterminous; we can produce numbers outside of these infinite sets.

We know that there are numbers s such that 1 < gcd(s, n) < s. These numbers s are "neutral" to n since they neither divide nor are coprime to n. For n = p prime, there are no neutral s < p, but we can show that s = p^e for e > 1 are neutral, since they cannot divide p, but also s (mod p) ≡ 0 and are not coprime to p. Further we can produce composite s from the smallest prime divisor p of p (p itself) and the smallest q coprime to p (the least prime that does not divide p; for any odd prime p, q = 2). This s = pq does not divide p since it is larger than p and for the factor q not one of p, and it cannot be coprime to p since gcd(pq, p) = p. Therefore we have produced two species of neutral s that apply to all numbers n, even for n = prime p that exceed p, but we have shown that there are neutral s < n for composite n > 4. We know that one species, those that divide a power of n, ascribe to regular R. The second species of neutral numbers does not ascribe either to R or to C, for the condition 1 < gcd(s, n) < s. We know that we can produce an infinite number of composite numbers that are the product of at least one prime p | n and q such that gcd(q, n) = 1. Therefore we can produce a third infinite tensor S distinct from the union of R and C. We know from the proofs related to neutral numbers that there are no other species and thus we can divide all positive integers into members of the three infinite tensors R, S, and C.

Now we can make one more set of observations.

It is known that the numbers t coprime to n in C are also coprime to any other number divisible by the distinct prime divisors p of n. We can extract the squarefree root of n, rad(n) = A007947(n), simply by multiplying the distinct prime divisors of n. For instance, rad(10) = 6, since 6 = 2 × 3, and 10 = 2² × 3. Therefore, the numbers t coprime to the dozen are also coprime to 6, and indeed to any other number that is the product of both 2 and 3 (i.e., numbers in A033845). Thus we can subscript C_6 since the numbers coprime to 6 are also coprime to (6, 10, 16, 20, 30, 40, 46, ...).

We know that another definition of regular number r is one such that all prime divisors p also divide n. Therefore it too pertains to A007947(n), and we can likewise subscript R_6. Since the tensors are coterminous (with the exception of the number 1 shared between R and C), we can subscript S_6. Indeed the semicoprime numbers s = {a, 12, 13, 18, 19, 1a, 22, 24, 26, 29, ...} are shared between any base 6 R_6 = 6 A003582.

Therefore, the RSC configuration for uncial is also shared by senary, octodecimal, tetravigesimal, etc., and that for decimal is shared with vigesimal, etc.

This means we can talk of shared efficiencies between bases with the same squarefree root. The regulars, semicoprimes, and coprimes of these bases are the same; the numbers whose fractions terminate, are mixed recurrent, and recurrent are shared, etc.
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SenaryThe12th
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Joined: 2:03 PM - Mar 01, 2018

3:45 AM - Aug 15, 2018 #2

What does "tensor" mean in this context?
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SenaryThe12th
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Joined: 2:03 PM - Mar 01, 2018

12:12 PM - Aug 15, 2018 #3

Thanks Sally.  What confused me is that Icarus does use the term "set" and "tensor" in the same sentence:
Example: the tensor C pertaining to uncial is {1, 5, 7, b, 11, 15, 17, 1b, 21, 25, 27, 2b, ...}. It is easy to see a pattern. C contains all numbers k such that k (mod n) ≡ t in T, where T is the (necessarily finite) set of totatives t of n (i.e., the reduced residue system of n).
seemingly drawing a distinction between C as a tensor and T as a set.  Arrays have more structure than sets do (e.g. they have an ordering)  but after staring at this post crosseyed I can't see if where Icarus uses the ordering....

Anyways, the best I can make out is that the conclusion is that Senary is the most kickass base:

Therefore, the RSC configuration for uncial is also shared by senary, octodecimal, tetravigesimal, etc., and that for decimal is shared with vigesimal, etc.
But that may just be my own particular prejudices causing me to read more into that than what's there :)
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icarus
Dozens Demigod
icarus
Dozens Demigod
Joined: 12:29 PM - Apr 11, 2006

2:39 PM - Aug 15, 2018 #4

I ought to have used the same word. Sally = wendy; she's been banned, I am sorry.
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jim
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jim
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Joined: 8:19 PM - Apr 20, 2012

10:23 AM - Aug 18, 2018 #5

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