We calculate $\displaystyle \frac{1}{1387} = 0.\overline{000000010111101000}_2$. Take the repeating part as a bitstring and rotate two steps to the right to obtain $000000000101111010$ with binary complement $101111111111010000_2 = 196560$.
The Leech lattice has kissing number $196560$.
Why should we expect a connection between the binary expansion of $\displaystyle\frac{1}{19\times73}$ and the kissing number of the Leech lattice?
The question asks for a mere hint simply because I see no reason why one could suspect that the orbit of $2$ mod $1387$ and the kissing number of a lattice should have anything to do with each other. In particular, one is a quantity, and the other is a sequence, that just happen to be represented by the same bitstring.
The answer that this is coincidental is deeply unsatisfying due to the following observation.
We examine other reciprocals that expand to an $18$-digit repeating part. Let $f(n)$ be the function taking $n$ to the the repeating part of $\displaystyle\frac{1}{n}$ in binary and denote by $\to^n$ the right-shift operation performed $n$ times. Then the following is $f$ evaluated on every integer $n$ where $2$ has order $18$ mod $n$ (that I found with sage), rotated to the closest number to $2^{16} \times 3 = 196608$ (we count rotations that start with $0$ as the binary complement). $$\begin{align} f(19 = 19) \to^2 &= 193158 = 2^{16}\times3 - 3450\\ f(27 = 3^3) \to^2 &= 194180 = 2^{16}\times3 - 2428\\ f(57 = 3^19) \to^2 &= 197757 = 2^{16}\times3 + 1149\\ f(133 = 7^119) \to^2 &= 197100 = 2^{16}\times3 + 492\\ f(171 = 3^219) \to^2 &= 196224 = 2^{16}\times3 - 384\\ f(189 = 3^37) \to^2 &= 196954 = 2^{16}\times3 + 346\\ f(219 = 3^173) \to^2 &= 196308 = 2^{16}\times3 - 300\\ f(399 = 3^17^119) \to^2 &= 196443 = 2^{16}\times3 - 165\\ f(513 = 3^319) \to^2 &= 196753 = 2^{16}\times3 + 145\\ f(657 = 3^273) \to^2 &= 196707 = 2^{16}\times3 + 99\\ f(1197 = 3^27^119) \to^2 &= 196662 = 2^{16}\times3 + 54\\ f(1387 = 19^173) \to^2 &= 196560 = 2^{16}\times3 - 48\\ f(1533 = 3^17^173) \to^2 &= 196650 = 2^{16}\times3 + 42\\ f(1971 = 3^373) \to^2 &= 196574 = 2^{16}\times3 - 34\\ f(3591 = 3^37^119) \to^2 &= 196589 = 2^{16}\times3 - 19\\ f(4161 = 3^119^173) \to^2 &= 196623 = 2^{16}\times3 + 16\\ f(4599 = 3^27^173) \to^2 &= 196593 = 2^{16}\times3 - 15\\ f(9709 = 7^119^173) \to^2 &= 196614 = 2^{16}\times3 + 6\\ f(12483 = 3^219^173^1) \to^2 &= 196602 = 2^{16}\times3 - 6\\ f(13797 = 3^37^173) \to^2 &= 196612 = 2^{16}\times3 + 4\\ f(29127 = 3^17^119^173) \to^2 &= 196605 = 2^{16}\times3 - 3\\ f(37449 = 3^319^173) \to^2 &= 196609 = 2^{16}\times3 + 1\\ f(87381 = 3^27^119^173) \to^2 &= 196608 = 2^{16}\times 3 \\ f(262143 = 3^37^119^173) \to^2 &= 196607 = 2^{16}\times3 - 1\\ \end{align}$$
Incidentally, $$f(11) \to^2 = 744$$