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May 16th, 2017, 06:02 AM   #1
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Conjecture about odd numbers

Any odd number O >=3 could be written at least once as :

O=pq - phi(pq)

where p and q are both odd primes (p<=q)
phi(pq) is the Euler totient of pq

Example :

O=3*3- phi(3*3)=9-phi(9)=9-6=3

Is this conjecture equivalent to GB?
Can we prove it?
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May 19th, 2017, 10:23 AM   #2
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72 views and still no comment.
I have found something new about the particular set of odd semi-prime numbers.
This set is like an infinite "forest" with infinite number of infinite "trees".
I created a concept similar to primality splitting all the odd prime numbers on 2 subsets.
An odd semiprime number is either a foot of tree (pied d`arbre in french) either a "leaf" (feuille).

9,15,25,33, and so on are feet of tree
21,39,55,57 and so on are leaves

A foot of tree is an odd semiprime number pq such as :
pq-phi(pq) is never a semiprime number

example : 129=3*43

129-phi(129)=129-84=45 (45 is not a semiprime number)

A lef is an odd semiprime number such as :
pq-phi(pq) is a semiprime number

example :

21-phi(21)=21-12=9 (9 is an odd semiprime number=3*3)


39-phi(39)=39-24=15 (15 is an odd semiprime =3*5)

This set have many specific properties.
For example : a square odd semiprime is ALWAYS a foot of tree
There are many theorems to find about this particular set of odd semiprime numbers.

Each tree have a height "h"
856687=13*65899 has a height h=7

856687 is a leaf
because 856687-phi(856687)=856687-790776=65911 which is a semiprime
If we compute consequently pq-phi(pq) we will have those numbers :


See below the details of pq-phi(pq)

pq, phi(pq),pq-phi(pq)

856687 790776 65911
65911 62424 3487
3487 3160 327
327 216 111
111 72 39
39 24 15
15 8 7
As you see 15 is the foot of the tree.
For the foot 15, there are leaves to build
15 will give 39 and 55
39 and 55 too will give more leaves and so on

The tree has only one foot but an infinite height with an infinite number of leaves.

Can you solve the tree starting by 15 with all the leaves at the height 7 ?
A program will do it quickly and will show if you understood what I presented above.

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May 19th, 2017, 01:36 PM   #3
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Originally Posted by mobel View Post
Any odd number $\displaystyle n \geq 3$ could be written at least once as :

$\displaystyle n=pq - \phi(pq)$

where $\displaystyle p$ and $\displaystyle q$ are both odd primes $\displaystyle (p \leq q)$
$\displaystyle \phi(pq)$ is the Euler totient of $\displaystyle pq$

Is this conjecture equivalent to GB?
Can we prove it?
Well what I know about odd numbers is that they are not even. However supposing that such values of $\displaystyle n,p,q$ exist as stated then we have

$\displaystyle n=pq - \phi(pq)$
$\displaystyle n=pq - (p-1)(q-1)$
$\displaystyle n=pq - (pq-p-q+1)$
$\displaystyle n=p+q-1$
$\displaystyle n+1=p+q$

This last line says that $\displaystyle n+1$ is any even number greater than or equal to 4 then it can be written at least once as a sum of two odd primes.

Well it is clearly not true for $\displaystyle n=3$ but past 3 it looks like Goldbach.
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May 20th, 2017, 05:53 AM   #4
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Let us start from the foot of the tree 15
height 1 : 15
height 2 : 39 55
height 3 : 111 319 391 159 559 703

15 will give us 2 leaves : 39 and 55

39 will give us : 111,319 and 391
55 will give us : 159,559 and 703

We could continue indefinitely and at each height we will generate more leaves
It is not easy to represent the tree as graphics maybe there are software doing such thing.

What will be interesting is to generate a sequence of feet of trees (height 1), leaves at height 2, height 3 and so on.
We could then treat them horizontally.
The goal is to locate graphically where the semi-prime numbers with some defined gap g=(q-p) are located.
My hope is to find some closed formula (or approximation) to say that some semi-prime number is at the height h and the gap g
I know that it is very hard to compute such formula but I still think that we could build step by step some rules (or theorems).
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May 20th, 2017, 06:33 AM   #5
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As the height is growing the number of leaves is exploding.
Excel is not helpful ...
For the foot of the tree = 15 height 4 I did not finished yet and I reached those 34 values :


So only a program could generate a tree with height > 7

I will continue digging without computing.
Goal : find some rules, some patterns.

Last edited by mobel; May 20th, 2017 at 06:53 AM.
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May 20th, 2017, 07:41 AM   #6
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Here are the first 300 odd semi-prime numbers called feet of tree :

9,15,25,33,35,49,51,65,77,87,91,95,115,119,121,123 ,129,143,161,169,177,183,185,187,205,209,213,215,2 17,219,221,237,247,259,287,289,291,295,303,309,321 ,329,335,341,355,361,371,377,395,403,407,411,427,4 37,447,453,469,473,485,489,493,505,511,515,519,527 ,529,533,537,545,551,565,573,579,581,583,591,611,6 29,635,655,669,671,679,681,687,707,713,717,721,723 ,731,745,749,767,779,781,793,799,803,807,813,815,8 17,831,835,841,843,849,851,869,871,899,917,923,933 ,939,943,955,961,965,979,993,1003,1007,1041,1047,1 057,1067,1099,1101,1115,1119,1133,1135,1139,1145,1 147,1149,1157,1159,1169,1177,1191,1195,1205,1211,1 219,1241,1247,1255,1257,1261,1263,1271,1285,1293,1 299,1313,1317,1333,1337,1345,1349,1351,1355,1357,1 363,1369,1371,1383,1385,1389,1397,1403,1405,1411,1 465,1469,1501,1507,1513,1529,1535,1537,1541,1555,1 561,1563,1565,1569,1577,1589,1591,1631,1641,1643,1 649,1651,1673,1681,1691,1707,1717,1727,1735,1739,1 745,1757,1763,1765,1769,1779,1781,1793,1795,1797,1 799,1803,1807,1817,1821,1829,1839,1841,1849,1851,1 853,1857,1883,1895,1897,1909,1915,1923,1929,1939,1 961,1963,1969,1977,1983,1985,1991,2005,2019,2021,2 033,2059,2071,2073,2077,2095,2105,2117,2119,2127,2 147,2149,2155,2167,2171,2177,2181,2195,2199,2201,2 209,2217,2227,2257,2263,2271,2285,2291,2305,2315,2 317,2319,2327,2329,2353,2359,2369,2395,2419,2427,2 429,2435,2449,2453,2455,2463,2469,2471,2479,2481,2 489,2491,2495

Remind : An odd semi-prime number f is such as : f phi(f) is never equal to another semi-prime number

For example :
9-phi(9)=9-4=5 (5 is not an odd semi-prime (it could be prime or product of more than 2 odd prime)

Hence showing that an odd semi-prime is a foot of tree or not is easy as long as we know its 2 factors.

Maybe if we study deeply their distribution we could reach some results.
Mapping those numbers is very important even if it is too hard.
Is every tree infinite or no?
Can we know the cardinal of the sub-set height h for any foot of tree?
Can we predict the height of any chosen odd semi-prime (RSA numbers for example)?
Are the RSA numbers foot or leaves?
and so on ....
Many questions and few answers
Lot of work is needed before reaching some useful results.
Good luck to those interested in discovering the first theorems.
It is easy to show that any odd prime number squared (p*p) is always a foot of tree.
There are 6076 out of the first 10.000 odd semi-prime numbers which are foot of tree.
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May 21st, 2017, 05:24 AM   #7
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Correction because I could not edit the post :

phi(9)=6 instead of 4
9-6=3 (3 is not an odd semi-prime number hence 9 is a foot of tree)

No one pointed out the mistake!!!

Thank you for not correcting me. It said it all about you!
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May 25th, 2017, 09:27 AM   #8
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I have found that the concept of "forest" could be expanded to sets other than the odd semi-prime numbers.
But the function : O=pq-phi(pq) is different.
In a "forest" there are infinite number of "trees" with infinite number of "feet of tree" and infinite number of "leaves". the height of each "tree" is infinite too.
In fact the concept of "forest" allow us to partition a set where each element is unique and all the elements cover entirely the set.
Now good luck to you.
I go back to my writings.
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May 26th, 2017, 06:35 AM   #9
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Here is a representation of one tree as example
Attached Images
File Type: jpg tree.jpg (19.3 KB, 6 views)
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May 26th, 2017, 07:01 AM   #10
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What is interesting is that you could could start from any leaf and find some leaves of the tree.
Example : 57
You need to factorize 57
hence 3+19=22
22-1=21 (21 is the leaf height k-1) where k is the height of 57
Now you could find all the leaves starting from 21 :
22=5+17 (both are prime) hence you obtain the leaf to the right 5*17=85
22=7+15 (15 is not prime)
22=11+11 (both are prime) hence you obtain 11*11=121
Now you know the leaves at the height k (you still do not know the height k)
We need to factorize 21
hence 3+7=10 which give us 9=10-1
If we test 9
9-phi(9)=9-6=3 (3 is not an odd semi-prime hence 9 is a foot of tree.

9 is at height 1
hence 21 is at height 2
and 57 at height 3

At first glance we could know without factorizing.
But if we go deep we will surely find some relations helping us to factorize.
I gave just few ideas on how to rebuild entirely the tree.
An algorithm + some rules will help us to compute all what we need.
I will let you give a try to any odd semi-prime odd.
For example : O=1633=23*71
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