Q18, A1. Surprised about midpoints

Posted on September 14, 2018

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I am extremely surprised by the following

observations about midpoints:

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Let a and b be positive integers such that a < b.

Let d = b – a.

Let n = ab.

Let s denote the square root of n.

Let F denote the floor of s.

Let C denote the ceiling of s.

Let M(a, b) denote the midpoint of the interval [a, b].

Let M(F, C) denote the midpoint of the interval [F, C].

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In a first draft, I have proved the following:

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When d = 1, we have M(a, b) = M(F, C).

When d = 2, we have M(a, b) = M(F, C) + 0.5.

When d = 3, we have M(a, b) = M(F, C).

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I am very surprised that M(a, b) – M(F, C)

first increases and then decreases.

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Example for d = 1.

a = 5, b = 6, n = 30, F = 5, C = 6,

M(a, b) = M(5, 6) = M(F, C).

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Example for d = 2.

a = 5, b = 7, n = 35, F = 5, C = 6,

M(a, b) = M(5, 7) = 6,

M(F, C) = M(5, 6) = 5.5.

=====================

Example for d = 3.

a = 5, b = 8, n = 40, F = 6, C = 7,

M(a, b) = M(5, 8) = 6.5,

M(F, C) = M(6, 7) = 6.5.

=====================

I will post a proof in a document as soon as possible.

======================

Note that M(a, b) is the arithmetic mean of a and b,

and s is the geometric mean of a and b.

It is well-known that when b > a, the arithmetic

mean is greater than the geometric mean:

https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

So, it is sure that we have M(a, b) > s.

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With best regards, Jean-Claude

Q18, A3. Positions of [a, b] for d = 1, 2, and 3.

Posted on September 15, 2018

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I thank Peter a lot for the answer 2.

As I said at the end of answer 1,

my results agree 100% with AM > GM

when n is not a perfect square

and AM = GM when n is perfect square.

Furthermore, my results agree 100% with Peter’s comments.

Peter’s comment about the related graph is important.

I have also wanted and still want

to insert graphs into my documents.

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I have finished the proofs that I promised yesterday.

They are posted in the document:

D32. Central pairs of complementary divisors of n

https://www.researchgate.net/publication/327558662_Central_pairs_of_complementary_divisors_of_n

=======================

My notation is the following:

Let n be a positive integer that is not a perfect square.

Let {a, b} denote a pair of complementary divisors of n,

with a < b, so that n = ab.

Let d = b – a = [The length of the interval [a, b]].

Let s denote the square root of n.

Let F denote the floor of s.

Let C denote the ceiling of s.

Let M(a, b) denote the midpoint of the interval [a, b].

Let M(F, C) denote the midpoint of the interval [F, C].

Since n is not a perfect square, we have

F < s < C, C = F + 1, and M(F, C) = F + 0.5.

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When d = 1, a = F, b = F + 1,

M(a, b) = F + 0.5 = M(F, C).

This is proved in Lemma 5 of D32.

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When d = 2, a = F, b = F + 2,

M(a, b) = F + 1 = M(F, C) + 0.5.

This is proved in Lemma 8 of D32.

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When d = 3, a = F – 1, b = F + 2,

M(a, b) = F + 0.5 = M(F, C).

This is proved in Lemma 9 of D32.

===========================

Note. The numbering of the results will

change every day, because I have a long

list of improvements to do everywhere.

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With best regards, Jean-Claude

Q18, A4. Positions of [a, b] for d = 1, 2, … , 12.

Posted on September 15, 2018

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I have extended my results of answer 3

from d = 1, 2, 3 to d = 1, 2, 3, …, 12.

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I simplify my notation in the following way.

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Let n be a positive integer.

We want to determine whether n has a pair (a, b)

of complementary divisors, with a < b,

that are at distance d from each other.

Complementary means n = ab.

At distance d from each other means b – a = d.

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For example, if n = 3670, the complementary of 10

is 367, because 3670 = (10)(367).

The distance between 367 and 10 is 367 – 10 = 357.

======================

We can immediately solve this problem

with the following method:

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Let s denote the square root of n.

For every real number x,

let F(x) = [The floor of x].

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Then the above problem is immediately

solved with

a = F(s) – F((d – 1)/2),

under the condition that a is greater than

or equal to a certain minimum a(d).

We just have to check whether with

this value of a we have a(a + d) = n.

=========================

The values of a(d) for d = 1, 2, 3, …, 12

are the following:

a(1) = 1

a(2) = 1

a(3) = 1

a(4) = 1

a(5) = 4

a(6) = 2

a(7) = 9

a(8) = 5

a(9) = 16

a(10) = 8

a(11) = 25

a(12) = 13

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I am constructing an example

that I will post in my next answer.

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With best regards, Jean-Claude

Q18, A5. Example n = 18,446,989 and d = 12.

Posted on September 15, 2018

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Let us apply my method of answer 4

to determine whether n = 18,446,989

has a pair (a, b) of complementary divisors

that are at distance d = 12 from each other.

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First, let us check that the condition that

the integer a is greater than

or equal to a(12) = 13 is satisfied:

If a = 13, then

b = a + d = 13 + 12 = 25,

and ab = (13)(25) = 325,

which is much smaller than n = 18,446,989.

Therefore, it is sure that

the condition a > a(12) is satisfied.

=======================

With the notation of answer 4,

s = SQRT(n) = 4294.9958,

F(s) = Floor(s) =4294,

F((d – 1)/2) = F((12 – 1)/2) = F(11/2) = F(5.5) = 5,

a = F(s) – F((d – 1)/2) = 4294 – 5 = 4289,

b = a + d = 4289 + 12 = 4301,

ab = (4289)(4301) = 18,446,989 = n.

Thus, n =18,446,989 possesses a pair of complementary

divisors that are at distance d from each other,

namely, the pair (4289, 4301).

=======================

I have not finished to write all the proofs.

I will post them as soon as possible.

======================

With best regards, Jean-Claude

Using the identity

4AB+(A-B)2 = (A+B)2 ,

all other results followed.

4AB= (A+B)2 - d2 and AB= ( A+B+d)/2 . (A+B-d)/2

The distance between ( A+B+d)/2 and (A+B-d)/2 is d.

I have tried to find solutions using the integer part of the square root (r), of the number n=AB for examble for 66, r=8 for 77, r=8 for 105, r=10 etc. So, if we say that S= 4r+3+N and N is an even number not smaller than -2 e.g. -2, 0, 2, 4, 6, etc then A and B can be given by the equations:

[S-sqrt(S^2 - 16n)]/2 and

[S+sqrt(S^2 - 16n)]/8

For examble for n=77=8^2+13 if we put 4 for N we have S=4.8+3+4=39 and

(S^2 - 16n) = 17^2, so (39-17)/2 = 11 and (39+17)/8 = 7

A possible research on the item should specify the parameters of N, e.g. for n=77 N=4 in terms of 13.

Q18, A8. Advantage: Only one square root.

Posted on September 16, 2018

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I thank Issam and Nikolaos a lot for the answers 6 and 7.

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The advantage of the method that I explained in the answer 4

is that if we have to try several distances d,

with my method, we compute only one square root,

namely, the square root of n.

===============

For example, for the following question:

What is the splitting P = AB of the product P of the first N primes that gives the smallest distance between A and B?

https://www.researchgate.net/post/What_is_the_splitting_P_AB_of_the_product_P_of_the_first_N_primes_that_gives_the_smallest_distance_between_A_and_B

The cases N = 1, 2, …, 7 are easy.

The case N = 8 is the first to be a little more difficult:

Given the product P of the first 8 primes

P = (2)(3)…(19) = 9,699,690,

find the factorization P = ab, with a < b

that gives the minimum distance d = b – a between a and b.

Because a and b do not have any common prime factor

and contain all the primes from 2 to 19, we have that

either d = 1

or d is a prime greater than 19,

or d is a product of primes greater than 19.

The smallest such product is (23)(23) = 529.

We have to try d = 1, 23, 29, 31, …, 523,

and only after d = 523, we have to worry about products of primes.

We compute only one square root:

s = [The square root of 9,699,690] = 3114.43… ,

and F(s) = [The floor of s] = 3114.

We try d = 23, 29, 31, and 37 without success.

Then we try d =41.

F((d – 1)/2) = [The floor of (d – 1)/2]

= F((41 – 1)/2) = F(40/2) = F(20) = 20.

a = F(s) – F((d – 1)/2) = 3114 – 20 = 3094

b = a + d = 3094 + 41 = 3135

ab = (3094)(3135) = 9,699,690 = P.

Thus, the minimum distance is d = 41,

and the corresponding factorization is P = (3094)(3135).

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With best regards, Jean-Claude

Q18, A9. Simplification of my method.

Posted on September 16, 2018

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Here is a simplification of my method of answer 4.

I still have to work on n_min.

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Let n be an integer greater than or equal to n_min.

Let d be a positive integer.

To determine whether there exists

an integer a such that a(a + d) = n

do the following:

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s = [The square root of n].

F = [The floor of s].

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Case 1: d is odd.

Let k = (d – 1)/2 so that d = 2k + 1.

Then the only possible value of a is a = F – k.

Let b = a + d.

Then the answer is yes if and only if ab =n.

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Case 2: d is even.

Let k = d/2 so that d = 2k.

Then the only possible value of a is a = F – k + 1.

Let b = a + d.

Then the answer is yes if and only if ab =n.

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For d = 1, 2, 3, and 4, n_min 1.

For d = 5, n_min = 25.

For d = 6, n_min = 8.

For d = 7, n_min = 121.

For d = 8, n_min = 49.

For d = 9, n_min = 361.

For d = 10, n_min = 15.

For d = 11, n_min = 841.

For d = 12, n_min = 120.

==================

Example n = 18,446,989 and d = 12.

The condition that n is greater

than or equal to 120 is satisfied.

s = 4294.9958,

F = 4294,

k = d/2 = 12/2 = 6,

a = F – k + 1 = 4294 – 6 + 1 = 4289.

b = a + d = 4289 + 12 = 4301.

ab = (4289)(4301) = 18,446,989 = n.

Thus, n =18,446,989 possesses a pair of complementary

divisors that are at distance 12 from each other,

namely, the pair (4289, 4301).

==================

With best regards, Jean-Claude

Q18, A10. Five big surprises.

Posted on September 16, 2018

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I recall my notation.

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Let n be a positive integer.

A pair of complementary divisors of n is a pair

{a, b} of positive integers such that n = ab.

For example, when n = 42, the pairs

{1, 42}, {2, 21}, {3, 14}, and {6, 7}

are the pairs of complementary divisors of 42.

A central pair of complementary divisors of n

(CPCD of n), is a pair {a, b} of complementary

divisors of n that are at the minimum distance

from each other.

For example, the pair {6, 7} is the central pair

of complementary divisors of 42.

I consider ordered pairs (a, b) of complementary

divisors of n such that a < b or a = b.

I consider the distance d = b – a between a and b.

From d = 1 to d = 12, I have found a positive

integer n_min(d) depending on d such that

for all n greater than or equal to n_min(d)

I have found a formula that gives me the pair

(a, b) of complementary divisors of n such that

b – a = d if such a pair exists.

For example, when n = 42,

there is the pair (6, 7) with d = 1

and no pairs with d = 2, 3, 4, …, 10.

Let s = [The square root of n].

F = [The floor of s],

C = [The ceiling of s],

M = [The midpoint of the interval [F, C] ],

that is: M = F + ½.

c = [The center of the interval [a, b] ],

r = [The radius of the interval [a, b] ] = d/2,

so that a = c – r and b = c + r.

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When my condition n > n_min(d) or n = n_min(d)

is satisfied, I have the following:

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First surprise: When d is odd, c = M

that is all the intervals [a, b] are centered at M.

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Second surprise: When d is even, c = C,

that is all the intervals [a, b] are centered at C.

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Third surprise: There no restriction on n

for d = 1, 2, 3, and 4:

In these four cases, n_min(d) = 1.

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Fourth surprise: The center c oscillates:

For d = 1, 2, 3, 4, 5, 6, ………

c = M, C, M, C, M, C, … .

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Fifth surprise: For every n and d satisfying

my condition, there is at most one odd

value of d and at most one even value of d

for which there exists a pair (a, b)

of complementary divisors of n

such that b – a = d.

Since my condition is satisfied for all n

when d = 1, 2, 3, 4, we have that

the cases d = 1 and d =3 exclude each other,

and the cases d = 2 and d = 4 exclude

each other.

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My new method is the following:

When d is odd, c = M = F + ½.

When d is even c = C = F + 1.

n = ab = (c – r)(c + r) = c^2 – r^2,

consequently

r^2 = c^2 – n,

r = [The square root of c^2 – n],

and we obtain the central pair {a, b} of

complementary divisors of n if this

square root is an integer:

a = c – r,

b = c + r.

======================

With this method, I immediately obtain

the CPCD of n when n is the product of

the first N primes, for N = 1 to N =9.

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I will post an example as soon as possible.

==================

I will write and post all the proofs in

my documents as soon as possible.

===================

With best regards, Jean-Claude

Q18, A11. Example: Product of the first 9 primes

Posted on September 16, 2018

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I apply my method posted in answer 10 above

when n is the product of the first 9 primes:

n = (2)(3)(5)(7)(11)(13)(17)(19)(23),

n = 223,092,870,

s = 14,936.29… ,

F = 14,936.

Since n contains only one even prime factor,

namely, the prime 2, and n = ab, this unique

even factor is either in a or in b, so that one

of the integer a and b is even, and the other

is odd. Therefore, d = a – b is odd.

Therefore,

c = M = F + ½ = 14,936.5.

c^2 = 223,099,032.25

c^2 – n = 6,162.25

r = [The square root of c^2 – n],

r = 78.5,

a = c – r = 14,936.5 – 78.5 = 14,858,

b = c + r = 14,936.5 + 78.5 = 15,015,

d = 2r = 157.

===============

Check:

a = 14,858 = (2)(17)(19)(23),

b = 15,015 = (3)(5)(7)(11)(13),

ab = [ 2)(17)(19)(23)] [(3)(5)(7)(11)(13)]

= (2)(3)(5)(7)(11)(13)(17)(19)(23) = n.

=====================

I will work on n_min(d) as soon as possible.

I will post all the proofs in my documents

as soon as possible.

===================

With best regards, Jean-Claude

Q18, A12. I have posted the start of a rewriting

September 24, 2018

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This is to tell all the contributors that I am very grateful

for all their contributions, and that I have not left the boat.

======================

I have started the beginning of the beginning of a huge

rewriting of everything that is in the links below, with

all the details, including the most microscopic details.

This very small start of the beginning is posted at the top

of the following document:

Central pairs of complementary divisors of n

https://www.researchgate.net/publication/327558662_Central_pairs_of_complementary_divisors_of_n

This document also contains its old version, which lies below the beginning of the rewriting.

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I have found much better results.

I have them on several drafts.

I have to rewrite them carefully.

They are not posted yet.

===================

All this is about the following:

https://www.researchgate.net/publication/327558662_Central_pairs_of_complementary_divisors_of_n

https://www.researchgate.net/publication/327392627_List_of_results_on_extensions_from_Euclid's_proof

https://www.researchgate.net/publication/327350053_Extensions_from_Euclid's_proof_of_the_infinitude_of_primes

https://www.researchgate.net/publication/326985440_Minimum_distance_from_splitting_the_product_of_the_first_N_primes

https://www.researchgate.net/publication/326971510_Splitting_in_Euclid's_proof_to_find_the_next_prime

https://www.researchgate.net/publication/327287317_Algorithm_about_extensions_of_Euclid's_proof

https://www.researchgate.net/post/Is_there_a_proof_that_the_fractional_part_of_the_square_root_of_a_pronic_number_is_increasing

https://www.researchgate.net/post/How_to_find_central_pairs_of_complementary_divisors_of_n

https://www.researchgate.net/post/What_is_the_splitting_P_AB_of_the_product_P_of_the_first_N_primes_that_gives_the_smallest_distance_between_A_and_B

https://www.researchgate.net/post/What_are_the_pairs_of_consecutive_positive_integers_whose_product_is_a_primorial

https://www.researchgate.net/post/Is_there_a_published_improvement_of_Euclids_Theorem

https://www.researchgate.net/post/What_are_the_products_nn_1n_2_equal_to_a_primorial

==================

With best regards, Jean-Claude