Cubes in a recursive succession

Number Theory Level 4

Let $(a_n)$ be a recursive relation with $a_n=a_{n-2}\times a_{n-1}$ , for all $n>2$ . If $a_1=18$ and $a_2=12$ , how many cubes will there be in the first 2016 terms?

The answer is 504.

1 solution

Arjen Vreugdenhil
Apr 27, 2016

We write $a_n = 2^{b_n} \cdot 3^{c_n}.$ Then the sequences $(a_n)$ and $(b_n)$ are Fibonacci-like sequences: $b_1 = 2,\ b_2 = 1,\ b_{n+2} = b_n + b_{n+1}; \\c_1 = 1,\ c_2 = 2,\ c_{n+2} = c_n + c_{n+1}.$ We need to count the indices $1 \leq i \leq 2016$ for which both $b_i$ and $c_i$ are multiples of 3. Now modulo 3, the sequence $(b_n)$ becomes $b_n \equiv -1, 1, 0, 1, 1, -1, 0, -1, -1, 1, 0, \dots\ \ \ \text{mod}\ 3,$ and for $(c_n)$ we find the same with opposite signs.

Thus we see that $b_i \equiv c_i \equiv 0$ mod 3 if and only if $i \equiv 3$ mod 4.

Therefore there are $2016/4 = \boxed{504}$ cubes in the original sequence $(a_n)$ .

Cubes in a recursive succession

The answer is 504.

1 solution

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