Let the sequence $b_n$ such that $$b_1=1, \; b_n=b_{n-1}+b_{\left[\frac{n}{2}\right]} \;\; \text{for all}\;\; n \ge 2,$$ where $\left[\frac{n}{2}\right]$ is an integer part of a real number $\frac{n}{2}$ ($\left[\frac{n}{2}\right]$ is a largest integer not exceeding $\frac{n}{2}$). Prove that the sequence $b_n$ contains infinitely many elements that are divisible by $7$.
My work. The sequence $b_n$ is $1, 2, 3, 5, \fbox{7}, 10, 13, 18, 23, 30, 37, 47, 57, \fbox{70}, 83, 101, \fbox{119}, 142, 165, 195, 225, 262, 299, 346, 393, 450, 507, 577, 647, 730, 813, 914, \fbox{1015}, \fbox{1134}, \fbox{1253}, 1395, 1537, 1702, 1867, 2062, 2257, 2482, 2707, 2969, 3231, 3530, \fbox{3829}, 4175, 4521, \fbox{4914}, 5307, 5757, 6207, 6714, 7221, \fbox{7798}, 8375, 9022, 9669, 10399, 11129, \fbox{11942}, 12755, 13669, 14583, 15598, 16613, 17747, 18881, 20134, 21387, 22782, 24177, 25714, \fbox{27251}, 28953, 30655, \fbox{32522}, 34389, 36451, 38513, 40770, 43027, 45509, 47991, 50698, 53405, 56374, 59343, 62574, 65805, \fbox{69335}, 72865, 76694, 80523, 84698, 88873, \fbox{93394}, 97915.$
It is not easy to see what is the pattern here.
First simplify to: $b_n=(b_{n-1}+b_{\left[\frac{n}{2}\right]}) \pmod 7$
Now the divisible by 7 test is just $=0$.
Observe that for each 0 that does occur, there is a sequence of three equal numbers. To be specific, if $b_n$ is 0, then $b_{2n}$, $b_{2n+1}$ and $b_{2n+2}$ are all the same. If those are 0 themselves, we have another match.
Observe that for each sequence of three equal non-zero numbers, we get a sequence of seven numbers in arithmetic progression (mod 7). To be specific, if $b_n$ is 0, for $m \in [0,6]$ we get $b_{4n+m} = (b_{4n}+m\times{b_{2n}}) \pmod7 $. One of them must be 0, so we have another match.
QED.