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 2008-12-03, 08:41 #1 devarajkandadai     May 2004 22·79 Posts 3-factor Carmichael numbers NECESSARY & SUFFICIENT CONDITIONS FOR A THREE-FACTOR COMPOSITE NUMBER WITH FOLLOWING SHAPE TO BE A CARMICHAEL NUMBER Let N, the composite number, have the shape (2m+1)(10m+1)(16m+1). Here m belongs to N. The necessary and sufficient conditions: a) (80m^2 + 53m + 7)/20 should be an integer b) The values of m which render the above an integer should also render 2m + 1, 10m + 1 and 16m + 1 prime. This is a corollary of the Devaraj-Pomerance-Maxal theorem (ref:: www.crorepatibaniye.com/failurefunctions
2008-12-03, 13:52   #2
davieddy

"Lucan"
Dec 2006
England

2·3·13·83 Posts

Quote:
 Originally Posted by devarajkandadai This is a corollary of the Devaraj-Pomerance-Maxal theorem (ref::
Pretentious moi?

 2008-12-03, 15:26 #3 alpertron     Aug 2002 Buenos Aires, Argentina 2·727 Posts I tested the first five million values of m and it appears to work. This is the program in UBASIC, noticing that the point "a" is equivalent to m=1 (mod 20). Nothing is printed, so it is OK. Code:  10 for M=1 to 5000000 20 if 2*M+1<>nxtprm(2*M) or 10*M+1<>nxtprm(10*M) or 16*M+1<>nxtprm(16*M) then 60 30 A=2*M+1:B=10*M+1:C=16*M+1:D=A*B*C 40 if (D-1)@(A-1)<>0 or (D-1)@(B-1)<>0 or (D-1)@(C-1)<>0 then K=0 else K=1 50 if (M@20=1 and K=0) or (M@20<>1 and K=1) then print M,(2*M+1)*(10*M+1)*(16*M+1) 60 next M Last fiddled with by alpertron on 2008-12-03 at 15:44 Reason: Added program in UBASIC
2008-12-03, 16:02   #4
R. Gerbicz

"Robert Gerbicz"
Oct 2005
Hungary

110001001012 Posts

Quote:
 Originally Posted by devarajkandadai NECESSARY & SUFFICIENT CONDITIONS FOR A THREE-FACTOR COMPOSITE NUMBER WITH FOLLOWING SHAPE TO BE A CARMICHAEL NUMBER Let N, the composite number, have the shape (2m+1)(10m+1)(16m+1). Here m belongs to N. The necessary and sufficient conditions: a) (80m^2 + 53m + 7)/20 should be an integer b) The values of m which render the above an integer should also render 2m + 1, 10m + 1 and 16m + 1 prime. This is a corollary of the Devaraj-Pomerance-Maxal theorem (ref:: www.crorepatibaniye.com/failurefunctions
This is trivial, because lcm(p1-1,p2-1,p3-1)=80m. n is Carmichael number if and only if n-1 is divisible by 80m, n is squarefree and odd number (this is the Korselt theorem), (n-1)/(80m)=(80*m^2 + 53*m + 7)/20

Why you don't write: (80m^2 + 53m + 7)%20==13m+7==13*(m-1)==0 mod 20 so the simple condition is that m-1 is divisble by 20.

Last fiddled with by R. Gerbicz on 2008-12-03 at 16:03

 2008-12-06, 04:16 #5 devarajkandadai     May 2004 22×79 Posts 3-factor C.Ns I wonder whether 561 is the only 3-factor C.N. with 3 as a factor . The reason is that if we were to keep 3 fixed and increase p_2 and p_3 indefinitely, k becomes asymptotic to 6. A.K.Devaraj

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