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.
from Hot Weekly Questions - Mathematics Stack Exchange
Peter
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