3-factor Carmichael numbers

[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

[davieddy]

Originally Posted by devarajkandadai

This is a corollary of the Devaraj-Pomerance-Maxal theorem (ref::

Pretentious moi?

[alpertron]

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.

   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

[R. Gerbicz]

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.

[devarajkandadai]

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