Super fibonacci

A super-Fibonacci sequence is determined entirely by its first two terms, $1$ and $x$ , say. The sequence then proceeds as

$1, x, (1+x), 2(1+x), 4(1+x), 8(1+x),\ldots$ .

Since adding all previous terms amounts to doubling the last term, all terms from the third onwards are of the form $2^k(1 + x)$ , for some $k \geq 0$ . If $2016 = 2^5 \times 63$ were one of these terms, then $k$ will be one of the $6$ values $k = 0,\ldots ,5$ , giving $6$ possible values for $x$ . These are $x = 62, 125, 251, 503, 1007, 2015$ . There is also the possibility that $x = 2016$ , so there are $\boxed{7}$ such sequences in total.

This problem was number 25 on the Intermediate Division for the 2016 Australian Mathematics Competition.