# Prove that this particular sequence contains an infinite number of sixes

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?

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Why doesn't it start with $2,7,1,4,4,1,6$? Where did the second $7$ in your listed sequence come from? – mjqxxxx Mar 3 '14 at 21:35
It seems the rule is to do every pair from the start one by one, not actually the two previous ones. – ploosu2 Mar 3 '14 at 21:40
sorry I was wrong, my English is terrible, however, the rule is: multiply any two numbers to get the next – user132839 Mar 3 '14 at 21:42
This is a past Putnam problem, right? Or maybe it's from the book "Putnam and Beyond"? – ahuff44 Mar 4 '14 at 4:10
@ahuff44 Don't know if it's a Putnam problem, but at oeis.org/A096381 there are reference to Loren C. Larson. This can be found at math.la.asu.edu/~ifulman/mat194/problem-solving.pdf and there it says it's from The Mathematics Student, Vol. 26, No. 2, November 1978. – ploosu2 Mar 4 '14 at 12:39

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).
How do you know $8,8,8$ appears in the sequence? By inspection? – msh210 Mar 4 '14 at 6:05