Consider an arithmetic sequence $\{11 + 13k : k\in\mathbb{N}\cup\{0\} \}$ Does this sequence contain five consecutive composites? If we look at some selections of five consec. elements: $$11, 24, 37, 50, 63 \quad 50, 63, 76, 89, 102 $$ one may get the feeling that Any selection of 5 consecutive elements always contains a prime.
EDIT: non-sensical gibberish
Or is there a counter-example to this claim?
On
No, because the density of primes in this, and any other, arithmetic progression is 0.
See https://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions for details.
On
A well known way of generating a sequence of $n$ consecutive numbers that aren't prime is to consider $(n+1)!+2, (n+1)!+3,\ldots, (n+1)!+n$ - it's quite obvious that each of these numbers are divisible by $2, 3,\ldots, n$.
We could perform that for $n=65$, the generated sequencemust contain five consecutive elements of your sequence, thus producing a counterexample (way larger than what has already been posted in a comment, but this way you can solve a lot of similar problems).
Consider the system of congruences $$13x\equiv -11\pmod{2}, \quad13(x+1)\equiv -11\pmod{3},\quad 13(x+2)\equiv -11\pmod{5},\quad 13(x+3)\equiv -11\pmod{7},\quad 13(x+4)\equiv -11\pmod{11}.$$ By the Chinese Remainder Theorem, this has infinitely many solutions. So there are infinitely many consecutive $5$-tuples of our sequence which are respectively divisible by $2,3,5,7,11$. The idea generalizes. We can get arbitrarily long strings of composites.