$10^n+1$ as a product of consecutive primes I am interested to know the examples or the systematic rule whether $10^n+1$ is a product of consecutive primes.
For example, 
when $n=1$,
we have $10^1+1=11$ is a product of a single prime 11. So, a True example.
When $n=2$,
we have $10^2+1=101$ is a product of a single prime 101. So, a True example.
When $n=3$,
we have $10^3+1=1001$ is a product of a product of consecutive primes $7*11*13$. So, a True example.
When $n=4$,
we have $10^4+1=10001$ is NOT a product of a product of consecutive primes $73*137$. So, a False example.
Are there more examples or the systematic rule whether $10^n+1$ is a product of consecutive primes?
 A: Using FactorInteger[] in mathematica you can try various values of n and see if they fit the pattern. I tried $n=21$ and $n=100$, neither one did.
(If you don't have mathematica, you can just enter "FactorInteger[number you want prime factorization of] into wolfram alpha).
A: I'm finding that $11$ divides $10^n+1$ for all odd $n$. $7*11*13$ divides $10^n+1$ for odd $n$ divisible by $3$. But $17$ is a divisor only if $n$ is an odd multiple of $8$. Thus it seems $10^n+1$ is not a product of consecutive primes for odd $n>3$.  If even $n$ is an odd multiple of $2$, $101$ divides $10^n+1$, but $103$ seems to be a divisor only for odd $n$ in sequence $17,119,221...$. If even $n$ is an odd multiple of $4$, $73$ and $137$ are consecutive divisors of $10^n+1$, but they are not consecutive primes. If even $n$ is an odd multiple of $8$, as already noted, $17$ is a divisor, but not $13$, and $19$ is a divisor only if $n$ is an odd multiple of $9$. Thus it seems, for $n>3$, $10^n+1$ could be the product of consecutive primes only if $16$ divides $n$. Is there a way to rule out these $n$ as well? For $n=16$ or any odd multiple of $16$, $10^n+1$ contains $353*449*641*1409*69857$, which are not consecutive primes. For $n=32$ and its odd multiples, $10^n+1$ contains $19841*976193*6187457*834427406578561$, again not consecutive primes. Thus if $10^n+1$ is not the product of consecutive primes when $n$ is a given power of $2$, then it is not the product of consecutive primes for any odd multiple of that power of $2$. This shortens the search among the even $n$ remaining. 
