Number of 1's in binary expansion of $a_n = \frac{2^{\varphi(3^n)}-1}{3^n}$ My question is about the Hamming Weight (or number of 1's in binary expansion) of $a_n = \frac{2^{\varphi(3^n)}-1}{3^n}$ A152007
For example, $a_3 = 9709 = (10110111101001)_2 $ has nine 1's in binary expansion
I guess the answer is $3^{(n-1)}$ but I can't prove it
Is that correct?

It is, indeed, correct. Notice first that $2-(-1)=3$ is divisible by $3$, so by lifting-the-exponent lemma the number
$$
A=\frac{2^{3^{n-1}}-(-1)^{3^{n-1}}}{3^n}=\frac{2^{3^{n-1}}+1}{3^n}
$$
is an integer. Notice also that for $n>0$ it has less than $3^{n-1}$ binary digits. Assume that it has $m$ binary digits. We have
$$
\frac{2^{\varphi(3^{n})}-1}{3^n}=\frac{2^{2\cdot 3^{n-1}}-1}{3^n}=A(2^{3^{n-1}}-1).
$$
Next, let $l=3^{n-1}-m$, $B=2^l-1$ and $C=2^m-A$. We claim that your number's binary expansion looks like a concatenation of $A-1$, $B$ and $C$, where we also add some zeros in expansion of $C$ until we get exactly $m$ digits. So, we should have
$$
a_n=C+2^mB+2^{m+l}(A-1).
$$
This equality is true, because
$$
C+2^mB+2^{m+l}(A-1)=2^m-A+2^m(2^l-1)+2^{m+l}(A-1)=
$$
$$
=(A-1)(2^{m+l}-1)+2^{m+l}-1=A(2^{m+l}-1)=A(2^{3^{n-1}}-1).
$$
Now, if the sum of digits of $A-1$ is equal to $s$, then the sum of digits of $C$ is $m-s$ and the sum of digits of $B$ is, of course, $l$, so the sum of digits of $a_n$ is
$$
s+m-s+l=m+l=3^{n-1},
$$
as needed.
For the particular case $n=3$, described in the post, $A=10011_2=19=\frac{2^9+1}{27}$ and $C=2^5-A=13=01101_2$ (notice the added zero)
