$3^k \cdot (n + \frac12) - \frac12$ powers of $2$

Question

Does the above sequence $3^k (n + 1/2) - 1/2$ always have a power of 2 in it for all n and some k?

Let $s_{nk} = 3^k (n + \frac12) - \frac12$, does $s_{nk} == 2^m$ for some m and k.

I'm thinking that because 3 is coprime with 2 that the sequence will eventually hit a power of 2 but it is a hunch and I can't seem to prove it.

Answer 1

Your equation reduces to

$$3^k(2n+1)=2^m+1$$

for some $k,n,m$.

If you're fixing $k$, take $m=3^{k-1}$. Then, by Lifting the Exponent Lemma (you can find a description and proof of the lemma along with notation here),

$$\nu_3\left(2^{3^{k-1}}+1\right)=\nu_3\left(2+1\right)+\nu_3\left(3^{k-1}\right)=k,$$

so

$$3^k\big|2^{3^{k-1}}+1\implies \frac{2^{3^{k-1}}+1}{3^k}=2n+1$$

for some $n$. So, fixing $k$, there always exists a pair $(m,n)$ for which $3^k(2n+1)=2^m+1$.

If you're fixing $n$, take $n=3$. We have that the left hand side is a multiple of $7$, while

$$2^m+1\in\{2,3,5\}\bmod 7,$$

so there are no integers $K$ for which

$$\frac{2^m+1}{2n+1}=K,$$

and certainly no powers of $3$.

Answer 2

$$ s_{nk}=3^k \cdot (n+ \frac12) - \frac12 \\ s_{nk}=3^k n + { 3^k-1\over 2} \\ $$

• For even $k$ we need even $n$ to have some power of $2$ in $s_{nk}$
• For odd $k$ we need odd $n$ to have some power of $2$ in $s_{nk}$