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Can a product $p_1\cdot\ \cdots\ \cdot p_n$ with consecutive primes $p_1,\cdots,p_n$ be a Carmichael number ?

I have used two different search strategies to find a possible example. The first is based on the smallest prime in the list. We can stop if in a product two primes $p,q$ occur with $p\mid q-1$ because then a Carmichael number is no longer possible.

This is the code (carm(v) is the self-defined function)

gp > carm(v)={p=prod(j=1,length(v),v[j])-1;select(m->Mod(p,m-1)==0,v)==v}
%1 = (v)->p=prod(j=1,length(v),v[j])-1;select(m->Mod(p,m-1)==0,v)==v
gp > forprime(r=3,5000,q=r;w=[q];n=q;gef=0;while(gef==0,q=nextprime(q+1);w=concat(w,q);n=n*q;if(carm(w)==1,print(v));if(gcd(eulerphi(n),n)<>1,gef=1)))
gp >

So, the smallest prime in the list must exceed $5\ 000$.

The second strategy is based on the number of factors. For $3$ factors for example, we have the code (with the function carm() from above) :

gp > [a,b,c]=primes(3);while(a<10^9,[a,b,c]=[b,c,nextprime(c+1)];if(carm([a,b,c])==1,print([a,b,c])))
gp >

So, for three factors, the smallest prime factor must exceed $10^9$. This can easily be extended.

But perhaps, someone has an easy proof that there cannot be a solution.

Peter
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    The formula $(6k+1)(12k+1)(18k+1)$ affords three factor Carmichael numbers when all three factors are prime. Bertrand's postulate assures that $6k+1$ and $12k+1$ are not consecutive primes. This mitigates against the occurrence of consecutive three factor cases, but does not prove that other three factor or larger instances cannot exist. – Keith Backman Nov 07 '20 at 17:05
  • The Carmichael numbers of that form are only a small subset of the Carmichael numbers with three prime factors, moreover, I allow arbitary many prime factors. – Peter Nov 08 '20 at 12:57
  • But for $3$ prime factors, we can use that , if the factors are $p,p+a,p+b$ ($0<a<b$), then $(a+1)(b+1)>p$ must hold and we should be able to find a heuristic clearly speaking against this if not even a refutation. – Peter Nov 08 '20 at 12:58

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