How often (i.e. asymptotic density) is the reversal of the binary representation of $7n$ is a multiple of $7$? If you reverse the binary digits of a multiple of $3$, the result is always a multiple of $3$. The same is not true for $7$, but it does appear to happen more often than $\frac{1}{7}$ of the time. For fun, I'll call a number $n$ such that the reverse of the binary digits of $7n$ represents a multiple of $7$ a "sevenly" number.
I wrote a simple Python program to compute the frequency of sevenly numbers, and after letting it run for an hour the proportion had been decreasing overall, but not very steadily, and it had yet to fall below $0.25$.
I was wondering if someone knew how to determine the asymptotic density of sevenly numbers. My intuition says it should be $\frac{1}{7}$, but the computations I've done seem to suggest it might not.
EDIT
My original program didn't tell me what number it was on, only the proportion, but I ran it for an hour, so it probably got into the millions. I've changed the program since then to print out the number it's on, too. Here is that code:
def brev(n):
    return int("0b"+(bin(n)[::-1])[:-2],2)

n = 1
sevenlies = 0
while True:
    if brev(7*n)%7 == 0:
        sevenlies += 1
    print("%d\t%f"%(n,sevenlies/n))
    n += 1

 A: Here is my reasoning on why
the number of times
the remainders mod $7$
match more than
randomly.
The key things,
to me,
are that
mod $7$,
$2^{3m} \equiv 1,
2^{3m+1} \equiv 2,
2^{3m+2} \equiv 4
$.
Therefore,
if
$m$ is one less than 
the number of bits in $n$,
write
$n = \sum_{k=0}^m b_k 2^k
$.
If $s(n)$ is the
value of $n$ mod $7$
and
$d(0, 1, 2) = (1, 2, 4)$,
then
$s(n) 
\equiv \sum_{k=0}^m b_k d(k \bmod 3)
$
Reversing the bits,
define
$r(n)
=\sum_{k=0}^m b_{m-k} 2^k
=\sum_{k=0}^m b_{k} 2^{m-k}
$.
Then, mod $7$,
$\begin{array}\\
s(r(n))
&=\sum_{k=0}^m b_{k} 2^{m-k}\\
&\equiv\sum_{k=0}^m b_{k} d(m-k \bmod 3)\\
\end{array}
$
so that
$\begin{array}\\
s(r(n))-s(n)
&\equiv\sum_{k=0}^m b_{k} (d(m-k \bmod 3)-d(k \bmod 3))\\
\end{array}
$
Whenever
$d(m-k \bmod 3)
=d(k \bmod 3)
$,
the term is zero.
This happens when
$m-k \equiv k \bmod 3$
or
$m \equiv 2k \bmod 3$
or
$2m \equiv k \bmod 3$.
This happens
about one third of the time
so that, 
independent of the value,
one third of the bits
will cancel out.
A: If the ratio converges to a limit, then it is $1/7$.
Let $n'$ be the reverse of $n$.
If $n$'s binary expansion is $\sum_{k=1}^m d_k 2^k$,
then, modulo $7$, we have
$n' = \sum_{k=1}^m d_k 2^{m-k} \equiv \sum_{k=1}^m d_k 2^{m-k} 8^k \equiv 2^m \sum_{k=1}^m d_k 4^k$,
and so $n'$ is a multiple of $7$ if and only if $\sum_{k=1}^m d_k 4^k \equiv 0$,
which means that $n$'s binary string, when read in base $4$, is a multiple of $7$.
Let $a_n$ be the number of sevenly numbers between $0$ and $2^n/7$, so the number of $n$-digits binary strings that are multiples of $7$ in both base $2$ and base $4$.
$(a_n)_{n \ge 0} = 1,1,1,2,3,5,10,16,28,56,\ldots$
If you let $a^{i,j}_n$ be the number of $n$-digits strings that represent, mmodulo $7$, $i$ when read in base $2$ and $j$ when read in base $4$, you have the recurrence relations $a^{i,j}_{n+1} = a^{4i,2j}_n + a^{4i+3,2j+5}_n$ and so if you put all those numbers in a vector $V_n \in \Bbb R^{49}$ there is a matrix $M$ such that $V_{n+1} = M V_n$.
The characteristic polynomial of $M$ is
$(x-2)(x^3-1)^4(x^9 - 10x^6 + 24x^3 - 8)^2(x^9 -3x^6 - 4x^3 - 1)^2$
$2$ is its largest eigenvalue, and its eigenspace is generated by the uniform vector $(1,1,1,\ldots,1)$, so as $n \to \infty$, the components of $V_n$ are all equivalent to $\frac 1 {49} 2^n$, and so as $n$ gets larger, $a_n = a_n^{0,0} \sim \frac 1 {49} 2^n$ which is $\frac 17$ of the $\frac 17 2^n$ numbers we are looking at.
