20190811, 05:01  #1 
May 2004
2^{2}×79 Posts 
Continued product Carmichael numbers
Let me give an example of a set of continued product Carmichael numbers:
a)2465 = 5*17*29 b)278545 = 5*17*29*113 c)93969665=5*17*29*113*337 d)63174284545 = 5*17*29*113*337*673 and e)169875651141505 = 5*17*29*113*337*673*2689 Algorithm for this type of c.p.Carmichael numbers is simple and I will illustrate how to derive b) above starting from a). Largest prime factor of a) is 29. Check the first prime generated by 28*k + 1; when k = 4 we get 113. 
20190813, 04:43  #2  
May 2004
13C_{16} Posts 
Quote:
b)172081= 7*13*31*61 c)31146661 = 7*13*31*61*181 d)16850343601= 7*13*31*61*181*541 Important point: possibility of constructing such spiral Carmichael numbers strengthens my conjecture that, r, the number of prime factors of a Carmichael number is not bounded. Last fiddled with by devarajkandadai on 20190813 at 04:50 

20190924, 03:14  #3  
May 2004
2^{2}·79 Posts 
Quote:
151813201 = 41*61*101*601 182327654401=41*61*101*601*1201 875355068779201 = 41*61*101*601*1201*4801* 12605988345489273601 = 41*61*101*601*1201*4801*14401 726117534688527648691201 = 41*61*101*601*1201*4801*14401*57601 

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