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?

  • 1
    $\begingroup$ Verified up to $n=17$ (CoCalc didn't let me check higher). It is a very intriguing result if it holds in general. $\endgroup$
    – Wojowu
    Jun 27, 2021 at 22:36

1 Answer 1


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)


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