# A few 'unproven' claims

The following properties/theorems (if we can call them that) were 'discovered' when investigating several topics that appear in the 'Number Theory' and 'Math' section of this website.
These have been verified empirically (as far a personal-use laptop can verify). The author (that's me) has reasonable confidence that these properties/conjectures, whatever we prefer to call them, are true. The reader is welcome to prove or contradict, either of which will be duly credited.

Why am I posting these here, rather than finding a proof for these myself, a journey that in itself would be enlightening? The reasons are several -

(1) Over the past several years, these were noted in random places (notepads, loose pages, etc.). Needed a reference list of the to-do backlog, though I may be missing several of them.
(2) Some claims may require know-how of concepts beyond my working knowledge. It is too tempting to discover new properties, than to spend time learning how to prove something that I believe to be reasonably true.
(3) In most instances, the impulse of moving ahead with the primary focus of the investigation was too compelling.
(4) To allow readers to provide inputs that may help with the proof, or give readers an opportunity to prove it themselves.

It is noted that the propositions in the list may seem trivial to an expert.
Claims 6 thro 10 are new findings related to "co-prime gaps" (See article on Co-prime Gaps). As I try to explain the theorem/claim, especially 6-10, it might seem a bit wordy. The formal statement of the claim, cited after the example, is concise.

Claim # 1:
Between every 2 natural numbers, there exists a real number (possibly irrational) such that the reciprocal of the number is the fractional part of the number. E.g. The reciprocal of golden ratio 1.618... is its fractional part 0.618... The page See article on Strange property of Golden Ratio gives a general formula for obtaining such numbers, and lists a few of them. Looks like there exists one between every 2 natural numbers!
(Can be proved using Fundamental theorem of Calculus, or Rolles Theorem ??)

Claim # 2:
If p is a prime, then phi(p !) = phi( (p-1) !) * (p-1)

Claim # 3:
See article on Clock hands meet
Imagine a closed curve and 2 points (or objects) starting at the same point on the curve, and moving along the perimeter at uniform, but different speeds.
Let the ratio of their speeds be whole number. The 2 objects will meet one or more times along the curve before meeting again at the starting point.
All the points where the 2 objects meet along the trajectory will divide the perimeter of the curve into equal parts.
Further, if there are n such objects, then the points where all n objects meet also divide the perimeter of the curve into equal parts.

Claim # 4:
For any arbitrary closed convex polygon/curve there exists at least one line passing throught the center of mass of the curve that divides the polygon into 2 equal parts. Extending the concept, there possibly exists a plane for any closed, convex 3 dimensional object such that the plane divides the object into 2 parts having equal volumes. So on...for higher dimensions.

Claim # 5:
When any 2 uniform masses are joined along their surfaces, that contains their center of mass, and if the center of mass of the resulting system remains in the plane of intersection, then the 2 masses are equal.

Claim # 6:
Here, we will be considering products of the form 3 x p2 x p3, or 5 x p2 x p3 x p4 x p5, or 7 x p2 x ... x p7, etc. where a prime number p1 (say 3, or 5 or 7, etc.) is multiplied with other prime numbers greater than p1, such that the product has only p1 terms (e.g 3 x 7 x 11, or 5 x 11 x 31 x 59 x 101, etc)
Now consider the sequence natural numbers from 1 to 3 x p2 x p3.
In this range of numbers, there are (3-1)!=2!=2 subsequences each having 4 consecutive numbers such that each number of this subsequence is a multiple of either 3 or p2 or p3 (i.e none of these 4 numbers are co-prime to the product 3 x p2 x p3.)
Eg. In the range 1 to 3 x 5 x 7, the 2 subsequences are {48,49,50,51} and {54,55,56,57} Similarly in the range 1 to 5x7x11x13x17, there are (5-1)!=24 subsequences having 6 consecutive numbers such that each number of this subsequence is a multiple of 5, 7, 11, 13 or 17.
Some of these are {1935, 1936, 1937, 1938, 1939, 1940}, {6135, 6136, 6137, 6138, 6139, 6140},... Similarly, a product starting with 7 contains 6! = 720 subsequences each of 8 consecutive numbers none of which are co-prime to the product.
For further discussion, let's call P as basic-odd-primordial (we are coining a new term here).
Also, let's call the subsequences as co-n-co subsequence. (consecutive non co-prime subsequence).
Thus, we can write our theorem as
Given an basic-odd-primordial of the form P = p1 x p2 x ... pk where the product contains p1 terms (i.e. k=p1), then the range of numbers from 1 to P consists of (p1-1)! co-n-co subsequences each of length (p1+1)
This is amazing because by means of multiples of primes (or by eliminating co-primes), the natural numbers are now themselves counting "factorials". (Sorry if you don't get this...can explain later :) ).

Claim # 7:
Referring to Claim # 6, if P consists of repeated instances of primes such that P = p1^k1 x p2^k2 x... then the number of subsequences is given by (p1-1)! x p1^(k1-1) x p2^(k2-1) x ...
E.g. For P = 3^2 x 5 x 7^101, the number of such co-n-co subsequences are (3-1)! x 3 x 7^100.

Claim # 8:
Referring to Claim # 6, if P' is an odd-primordial such that P' = P (i.e. basic-odd-primordial) x p(k+1) x p(k+2), etc., then number of co-n-co subsequences in the range 1 to P' is a multiple of number of co-n-co subsequences in the range 1 to P.
TODO: Exact formula yet to be derived.
E.g. For P' = 3 x 5 x 7 x 11, the number of subsequences = 20 which is a multiple of 2!
And, For P' = 5 x 7 x 11 x 13 x 17 x 19, the number of subsequences = 1368 which is a multiple of 4!=24

Claim # 9:
This one may be a bit difficult to understand, but will add more explanation later. It is interesting however, as it gives us a peek into strange but exotic nature of numbers, their muliples, & co-primes.
If we list all the co-n-co subsequences by removing the first and last terms of each, then the resulting list exhibits all permutations containing p2, p3,...pk such that
[ first number is multiple of p2, second is multiple of p3, ... ]
[ first number is multiple of p3, second is multiple of p2, ... ]
...
For a basic-odd-primordial, every permutation occurs only once (this is cool!)

To visualize this, consider for P = 3 x 5 x 7, the co-n-co subsequences are {48,49,50,51} and {54,55,56,57}.
Looking at 49 & 50 we see that 49 is a multiple of 7 and 50 is a multiple of 5,
whereas, for 55 & 56, these are multiples of 5 and 7.
Thus, we have co-n-co with multiples of [5,7] and then also the mutiples of permutation [7,5]. Thus each permutation occurs only once.
Similar pattern can be observed for P = 5 x p2 x p3 x p4 x p5, where the co-n-co subsequences contain every permutation consisting of multiples of
(p2, p3, p4, p5), (p2, p3, p5, p4), and so on (which are 4!=24 in all). Each permutation occurs only once.
Why does 'nature' (of numbers) have to play out every permutation? The proof, I guess, will answer this question.

Claim # 10:
If we have multiples of type P' = P (i.e. basic-odd-primordial) x p(k+1) x p(k+2), etc., then too, Claim # 9 holds true.
E.g. In case of P' = 3 x 5 x 7 x 11 x 19, we see that co-n-co subsequences will exhibit all possible permutations such that these are multiples of
[5,7,11], [5,7,19], [5,11,19] and will also include multiples of [5x7,11,19], [5,11x7,19], [5,11,19x7], so on.

Claim # 11:
Given p1 and p2 as two distinct primes, assuming p1 < p2. Consider the expression p1 x k = r (mod p2). As k varies from 1 to p2, r takes all values from 0 to p2-1 (not necessarily in that order).
That is, remainder r takes p2 distinct values from 0 to p2-1.

Example: Consider 3 and 7 to be p1 and p2. We will find remainders of 3 x k modulo 7 as k varies from 1 to 7. Thus we find
3 x 1 = 3 (mod 7) , 3 x 2 = 6 (mod 7) , 3 x 3 = 2 (mod 7) , 3 x 4 = 5 (mod 7) , 3 x 5 = 1 (mod 7) , 3 x 6 = 4 (mod 7) , 3 x 7 = 0 (mod 7).
Thus the remainders are 3, 6, 2, 5, 1, 4, 0 which are all values from 0 - 6.
This works for any 2 primes.

Multiple Claims
See article on Co-prime Gaps
Several properties mentioned here could use proofs.

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