Choose three arbitrary positive integers $\displaystyle u_1<u_2<u_3$ such that any two are coprime.

Define:

$\displaystyle L=lcm(u_1,u_2,u_3)= u_1u_2u_3$

$\displaystyle C=u_2u_3+u_1u_3+u_1u_2$

$\displaystyle D=u_1+u_2+u_3$

Lemma: $\displaystyle gcd(L,C)=1$ See proof here: http://www.mathhelpforum.com/math-he...tml#post303617

Let $\displaystyle a=L$ and $\displaystyle b=(-D)(C^{-1}) (\bmod L)$

Note that the above lemma is a necessary and sufficient condition for $\displaystyle C^{-1}$ to exist.

Define:

$\displaystyle p_1=u_1(am+b)+1$

$\displaystyle p_2=u_2(am+b)+1$

$\displaystyle p_3=u_3(am+b)+1$

Theorem: For any m such that $\displaystyle p_1,p_2,p_3$ are all prime, $\displaystyle n=p_1p_2p_3$ is a Carmichael Number.

Proof: Assume $\displaystyle p_1,p_2,p_3$ are all prime. For their product to be a Carmichael Number, it is sufficient to show that $\displaystyle lcm(p_1-1,p_2-1,p_3-1)|n-1$

Substituting via the above construction,

$\displaystyle n-1=( u_1(am+b)+1)( u_2(am+b)+1)( u_3(am+b)+1)-1$

$\displaystyle n-1= u_1u_2u_3(am+b)^3+(u_2u_3+u_1u_3+u_1u_2)(am+b)^2+( u_1+u_2+u_3)(am+b)$

$\displaystyle n-1= L(am+b)^3+C(am+b)^2+D(am+b)$

Since $\displaystyle p_i-1= u_i(am+b)$ , each of these divide the cubed term, and $\displaystyle am+b$ divides all terms.

Thus, the sufficient condition becomes: $\displaystyle u_i| C(am+b)+D$ for all $\displaystyle i$

Alternatively, $\displaystyle u_1u_2u_3|C(am+b)+D$

Or, $\displaystyle C(am+b)+D \equiv 0 (\bmod L)$

$\displaystyle (Ca)m+(Cb+D) \equiv 0 (\bmod L)$

Notice that by construction, $\displaystyle a=L$ and $\displaystyle b=(-D)(C^{-1}) (\bmod L)$, therefore this condition is true for all $\displaystyle m$.

Also notice that for this to hold for all $\displaystyle m$, $\displaystyle L|Ca$ and $\displaystyle L|Cb+D$. Therefore this choice for $\displaystyle a,b$ is unique.

Q.E.D

We now have a method, given an arbitrary pairwise coprime triplet $\displaystyle u_1<u_2<u_3$ , of generating a string of Carmichael Numbers!

Results:

For $\displaystyle u=(1,2,3), a=6, b=0 : (6k+1)(12k+1)(18k+1)$ generates Carmichaels 1729, 294409, 56052361, 118901521 …

For $\displaystyle u=(1,3,5), a=15, b=12 : (15k+13)(45k+37)(75k+61)$ generates Carmichaels 29341, 1152271, 64377911, 775368901 … (For this one, k must be even, so a=30 works as a refinement)

For $\displaystyle u=(1,4,7), a=28, b=4 : (28k+5)(112k+17)(196k+29)$ generates Carmichaels 2465, 6189121, 19384289, 34153717249 …

For $\displaystyle u=(3,11,20), a=660, b=2 : (1980k+7)(7260k+23)(13200k+41)$ generates Carmichaels 6601, 191614761361, 5124854089205401 ...