- **Math**
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- - **3-factor Carmichael numbers**
(*https://www.mersenneforum.org/showthread.php?t=11079*)

3-factor Carmichael numbersNECESSARY & 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:: [url]www.crorepatibaniye.com/failurefunctions[/url] |

[quote=devarajkandadai;151758]
This is a corollary of the Devaraj-Pomerance-Maxal theorem (ref:: [/quote] Pretentious moi? |

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 [/code] |

[QUOTE=devarajkandadai;151758]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:: [url]www.crorepatibaniye.com/failurefunctions[/url][/QUOTE] 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. |

3-factor C.NsI 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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