Dubs theory thread

WELCOME DUBS THEORY RESEARCHERS

here you can talk about your most recent finds regarding dubs theory, I will start with the well know ``dubsless primes'':

- let к(n) be the number of bases in which n has dubs excluding base 1 and base n-1 as these are trivial, we shall call n a dubsless number if к(n)=0 and n>3

- the dubsless numbers up to 10000 are 5, 6, 29, 41, 61, 113, 317, 509, 569, 761, 797, 1213, 1229, 2617, 5297, 6221 and 8017. it turns out that all of these numbers except 6 are prime, and up to 10 million all except 6 are prime, we call these primes the ``dubsless primes'' a new kind of primes. this raises the following questions in dubs theory:

- is the set of dubsless numbers/primes infinite?
- is 6 the only non prime dubsless number?

is 6 the only non prime dubsless number?

I didn't spend a lot of time writing and reviewing this, so there might be some errors or strange parts:

DEFINITIONS:

nⒹ Means the number n has dubs in some non-trivial base (any base other than 1 or n-1).

nⒹb Means the number n has dubs in base b.

nⒹ_d Means the number n has dubs ending in digit d in some base.

nⒹ_db Means the number n has dubs in base b, and the dubs digits are d.

Concrete example: 35Ⓓ₅6 Means 35 in base 6 has dubs ending in digit 5.

THEOREMS:

The square dubs theorem: ∀x∈ℕ(x²Ⓓ₀x) or every squared natural number n² represented in base n has dubs ending in digit 0. The proof of this simple theorem is left to the reader.

The first d-digit dubs theorem: (d+1)²-1 is the smallest number n which verifies nⒹ_d, the base that verifies that is d+1. Also n is a multiple of d. For example for digit 1, the first number that has dubs ending in 1, is (1+1)²-1=3 (11 in base 2). The proof of this simple theorem is also left to the reader.

The little composite dubs theorem: if we add integer multiples of d (d*i) to (d+1)²-1 (last theorem), the resulting number n = (d+1)²-1+d*i still verifies nⒹ_d, concretely nⒹ_dd+1+i. Also n continues to be a multiple of d. In other words, every multiple of d equal or greater than (d+1)²-1 has dubs in some base. Then, if a composite number c has a prime factor p for which c>=(p+1)²-1, c must have dubs.

The composite dubs theorem: If there were any other dubsless composite numbers c other than 6 they should be in some interval between (p_i+1)²-1 and (p_i+1+1)²-1 (both dubs, because last theorem), where p_i and p_i+1 are two consecutive primes. The lower bound means that if c has a prime factor p less or equal than p_i, then c must has dubs (last theorem), and the upper bound can be rewritten as p_i+1*(p_i+1+2), which is the product of two consecutive odd numbers (in some cases both may be prime). With those restrictions, there are 3 choices for the prime factors of c:

- greater than p_i+1²: p_i+1*(p_i+1+2) this only works when (p_i+1+2) is also a prime number and is the only choice because 2 is the minimum distance between two consecutive primes (excluding 2 and 3, we treat the interval defined by these two primes as a special case). But because last theorem shows us that the rewritten form of p_i+1*(p_i+1+2) has dubs, this choice doesn't give dubsless numbers.

- equal to p_i+1²: because the square dubs theorem this choice neither gives dubsless numbers.

- less than p_i+1²: then some prime factor p has to be less or equal than p_i, so because of the lower bound, this choice also gives dubs numbers.

This proves that no other composite number than 6 can be a dubsless number. And 6 is dubsless because the distance between the primes 2 and 3 is only 1, which doesn't happen for any other pair of primes.