Is there an 11-term arithmetic progression of primes beginning with 11?

Such an integer $C$ exists. The smallest $C$ with this property is $C=1536160080$.

I found this $C$ by computing the analogous number $C$ for a $3$-term prime arithmetic progression beginning with $3$, a $5$-term prime arithmetic progression beginning with $5$ and a $7$-term prime arithmetic progression beginning with $7$. This gave me the numbers $2,6,150$. When I plugged these into OEIS I found that the next term in this sequence is $1536160080$. You can see the relevant OEIS page here.

Siemion Fajtlowicz has been promoting the topic of $p$-long arithmetic progressions of primes which start with $p$ during 1993-4 (or longer). He and his colleague Micha Hofri got an $11$-long progression. Then, soon after, I got a small theorem which allowed me to get bunches of such $11$-progressions very fast, and also a lot of $13$-progressions (of $13$ primes starting with $13$) almost as quickly.

On the other hand, I conjectured that only a finite number of primes $p$ start $p$-progressions. Moreover, I believe that no prime $p>13$ starts any. Perhaps there is already none for $\ p=17$. (I got my results during 1994).

See also: http://primerecords.dk/aprecords.htm

For the sake of easy education let me mention the first simplest step toward finding $p$-long arithmetic progressions of primes which start with $p$.

Let $q$ be an arbitrary prime. Then the arithmetic progressions of more than $q$ integers must have a term divisible by $q$ when the difference of the progression is not divisible by $q$.

Now, let $\ p_0<p_1<\ldots\ $ be the sequence of all primes. Let there be a $p_n$-progression as described in this thread (in the first sentence of this answer/post). Then the difference of the progression must be divisible by the product of the previous primes, i.e. by

$$ \prod_{k<n}p_k. $$

Now, it is enough to check consecutive multiples of this product as the possible differences of the required progression.

Today, this remark would suffice toward a fast computation of the 11-progressions but around y.1992 it would make a computing station sweat for long hours (or days). But even in those days, the next step (a small theorem) was already good enough for computing even 13-progressions (but not 17-progressions).