Given the sequence
$$2,7,1,4,7,4,2,8,\ldots$$
which begins with $2, 7$ and is constructed by multiplying successive pairs of its members and adjoining the results as the next one or two members of the sequence. Prove that this sequence contains an infinite number of sixes.
Any idea?
For any block of consecutive elements of the sequence, the products of consecutive pairs form another block that will appear somewhere further on. The following blocks form a cycle in this manner: $$8, 8, 8 \to 6, 4, 6, 4 \to 2, 4, 2, 4, 2, 4 \to 8, 8, 8, 8, 8$$
Since $8, 8, 8$ appears in the sequence starting at position 72, there will be an infinite number of $6$s in the sequence (as well as $2$s, $4$s, and $8$s).