May 16th, 2017, 07:02 AM  #1 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41  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 O=3*3 phi(3*3)=9phi(9)=96=3 O=7 O=3*5phi(3*5)=15phi(15)=158=7 O=5 O=25phi(25)=5 etc... Is this conjecture equivalent to GB? Can we prove it? 
May 19th, 2017, 11:23 AM  #2 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
72 views and still no comment. I have found something new about the particular set of odd semiprime 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 : pqphi(pq) is never a semiprime number example : 129=3*43 129phi(129)=12984=45 (45 is not a semiprime number) A lef is an odd semiprime number such as : pqphi(pq) is a semiprime number example : 21=3*7 21phi(21)=2112=9 (9 is an odd semiprime number=3*3) 39=3*13 39phi(39)=3924=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 856687phi(856687)=856687790776=65911 which is a semiprime If we compute consequently pqphi(pq) we will have those numbers : 856687,65911,3487,327,111,39,15 See below the details of pqphi(pq) pq, phi(pq),pqphi(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. Thank you. 
May 19th, 2017, 02:36 PM  #3  
Senior Member Joined: Feb 2010 Posts: 637 Thanks: 106  Quote:
$\displaystyle n=pq  \phi(pq)$ $\displaystyle n=pq  (p1)(q1)$ $\displaystyle n=pq  (pqpq+1)$ $\displaystyle n=p+q1$ $\displaystyle n+1=p+q$ 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.  
May 20th, 2017, 06:53 AM  #4 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
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 semiprime numbers with some defined gap g=(qp) are located. My hope is to find some closed formula (or approximation) to say that some semiprime 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). 
May 20th, 2017, 07:33 AM  #5 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
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 : 327 471 535 951 1111 1167 1639 1671 2047 2103 2191 2407 2911 3127 3151 3799 3991 4927 5311 5671 5959 6319 7087 7111 8983 10279 10471 11911 15007 19351 20191 20863 28423 30439 39223 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 07:53 AM. 
May 20th, 2017, 08:41 AM  #6 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
Here are the first 300 odd semiprime 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 semiprime number f is such as : f phi(f) is never equal to another semiprime number For example : f=9=3*3 9phi(9)=94=5 (5 is not an odd semiprime (it could be prime or product of more than 2 odd prime) Hence showing that an odd semiprime 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 subset height h for any foot of tree? Can we predict the height of any chosen odd semiprime (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 semiprime numbers which are foot of tree. 
May 21st, 2017, 06:24 AM  #7 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
Correction because I could not edit the post : phi(9)=6 instead of 4 96=3 (3 is not an odd semiprime 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! 
May 25th, 2017, 10:27 AM  #8 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
I have found that the concept of "forest" could be expanded to sets other than the odd semiprime numbers. But the function : O=pqphi(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. 
May 26th, 2017, 07:35 AM  #9 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
Here is a representation of one tree as example

May 26th, 2017, 08:01 AM  #10 
Banned Camp Joined: Dec 2013 Posts: 1,117 Thanks: 41 
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 57=3*19 hence 3+19=22 221=21 (21 is the leaf height k1) 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 21=3*7 hence 3+7=10 which give us 9=101 If we test 9 9phi(9)=96=3 (3 is not an odd semiprime 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 semiprime odd. For example : O=1633=23*71 

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conjecture, numbers, odd 
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