Is there always one integer between these two rational numbers?

Question

It appears that for each integer $k\geq2$, there is always one integer $c$ that satisfies the inequalities below. Can this be proved?

$$\frac{3^k-2^k}{2^k-1}<c\leq \frac{3^k-1}{2^k}.$$

Note that for $k\geq2$ the lower bound is always a proper fraction and will never match an integer.

Edit 28/1/2018 I have a short proof here on Overleaf. <--- It's done! Edit #28 is perfect!

edit It looks like Waring's problem already has a solution. The sequence: https://oeis.org/A060692 "Corresponds to the only solution of the Diophantine equation 3^n = x*2^n + y*1^n with constraint 0 <= y < 2^n." This is equivalent to our $a+c$ for the Waring if statement. Oops, there is no proof that the diophantine equation is always $\leq 2^n.$ Here is a proof sketch of Waring's

Here is a brute force function:
aplusc[k_] := Module[{c}, c = 1; While[0 < 3^k - 2^k (++c)]; 3^k - 2^k (--c) + c]
where we increment $c$ until the calculation becomes negative, then we decrement by one to get $c$. We recalculate using that $c$ and add them together. A060692(k) equals this value.

Closed form: here. The brute force function illustrates the sawtooth pattern.

We can also create a(k), b(k), and c(k) using the same module. And we can use b(k) for the proof. Trivially, we can show b$(k) + 1 \leq 2^k.$

edit. Some more information and references: https://oeis.org/A002379 PM

Answer 1

It seems to me, that a sufficient proof for your first inequality is given in Zudilin,W., A new lower bound for ||(3/2)^k|| (manuscript at 2005$\,^\dagger$).

There he refers to a proof, where your $0.5^k$ (extracted from your ${3^k - 1\over 2^k}$ by writing ${3^k \over 2^k}-{1\over 2^k}=1.5^k-0.5^k$) was even replaced by the larger value $0.577^k$ . This means, that the integer value $c = \lfloor 1.5^k \rfloor$ is proven smaller than $ 1.5^k-0.577^k$. (The value $0.75^k$ in the Waring-conjecture however is still out of reach)

The left expression in the lhs inequality in your concatenated inequality is larger than rhs$-1$ (obvious by rewriting ${3^k-2^k\over 2^k-1}={3^k-1\over 2^k-1}-1 $ and then by expanding the geometric series) and the distance between the lhs and the rhs tends with increasing $k$ quickly towards $1$ and thus at most one integer value can be in the interval between lhs and rhs (namely the value $c=\lfloor (3/2)^k \rfloor $).

So I assume that with some additional work your full (concatenated) inequality might be provable with elementary means.

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$\,^\dagger$ Preprint for "Journal de Theorie des Nombres de Bordeaux", it is available on Zudilin's homepage here.

A view into S. Finch's book "mathematical constants" gives some quick insight, see this link to google-books:

Answer 2

To show $k\geq2$ per the comment in equality (3) on the other worksheet, we craft a logical expression which contains all the conditions: $\text{lexp }=k\geq 2\land 1\leq a<2^k\land 1\leq b<2^k-1\land \frac{3^k-1}{2^k-1}-\frac{b}{2^k-1}=\left(\frac{3}{2}\right)^k-a\ 2^{-k}=c,$ where $a, b$ must be in their respective bounds or we do not have fractional parts and the final condition defines the common floor.

Using Mathematica, we reduce lexp using the six different solution patterns: $\text{case }= \{\{a, b, c\}, \{a, c, b\}, \{b, a, c\}, \{b, c, a\}, \{c, a, b\}, \{c, b, a\}\}$ Table[{LogicalExpand[Reduce[lexp, case[[n]], Reals]]}, {n, 1, Length[case]}] $\begin{cases} 1&b=2^{-k} \left(2^k a-a-2^k+3^k\right)\land c=2^{-k} \left(3^k-a\right)\land k\geq 2\land a<\frac{2^{2 k}-3^k}{-1+2^k}\land 1\leq a\\ 2&b=2^{-k} \left(2^k a-a-2^k+3^k\right)\land c=2^{-k} \left(3^k-a\right)\land k\geq 2\land a<\frac{2^{2 k}-3^k}{-1+2^k}\land 1\leq a\\ 3&\frac{2^k b+2^k-3^k}{-1+2^k}=a\land c=2^{-k} \left(3^k-a\right)\land k\geq 2\land b<-1+2^k\land 2^{-k} \left(-1+3^k\right)\leq b\\ 4&c=\frac{-b+3^k-1}{-1+2^k}\land 3^k-2^k c=a\land k\geq 2\land b<-1+2^k\land 2^{-k} \left(-1+3^k\right)\leq b\\ 5&b=2^{-k} \left(2^k a-a-2^k+3^k\right)\land 3^k-2^k c=a\land k\geq 2\land \frac{-2^k+3^k}{-1+2^k}<c\leq 2^{-k} \left(-1+3^k\right)\\ 6&3^k-2^k c=a\land -2^k c+c+3^k-1=b\land k\geq 2\land \frac{-2^k+3^k}{-1+2^k}<c\leq 2^{-k} \left(-1+3^k\right)\\ \end{cases}$ Since these cases are the only possible solutions, and since each case contains $k\geq2$, we can state, "All six cases produce identical values for $a, b, c$, iff $k\geq2,$ as required."$\square$
29 Sep, 2017
We have enough information from the cases above to solve Waring's problem. First, we extract three boundaries and explain their formulas:
1) Upper bound of the numerator $a$ of the fractional part, case(1), $a<\frac{4^k-3^k}{2^k-1} = 2^k (1-\delta (k)).$ This boundary increases proportionally to the decrease of $\delta (k).$
2) Lower bound of the common floor $c,$ case(5), $c>\frac{3^k-2^k} {2^k-1} = 2^k \delta (k).$
3) Upper bound of the common floor $c,$ case(5), $c\leq\frac{3^k-1} {2^k} = \frac{\delta (k)}{\frac{1}{3^k-1}+\frac{1}{2^k-1}}.$
Note: $\left\lceil 2^k \delta (k)\right\rceil=\left\lfloor \frac{\delta (k)}{\frac{1}{3^k-1}+\frac{1}{2^k-1}}\right\rfloor.$

From: Waring's problem,
$g(k)=2^k+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor-2\ \ \ \ \text{ if }2^k \left\lbrace\left( \frac{3}{2} \right)^k \right\rbrace+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor\leq 2^k$,
where $\{\cdot\}$ is the fractional part.
Inspecting the "if" statement, we see that the product isolates the numerator of the fractional part, so we substitute $a$ and then substitute $c$ for the floor to get: $a+c\leq 2^k$. Empirically, this is solid. Note: $a+c$ is OEIS sequence A060692.
Next, we substitute the upper boundary for $a$ and the lower boundary for $c$ and change to an equality: $ 2^k (1-\delta (k)) + 2^k\delta (k) =2^k\iff k\geq1\land k \in \mathbb{Z}.$ This is true because $(1-\delta (k))$ and $\delta (k)$ are proportions-of-the-whole, which retain the proportionality when multiplied by the same value; which affirms that the boundaries are rigid (and that the sum will not be greater than $2^k$).

Empirical: $\left\lfloor 2^k (1-\delta (k))\right\rfloor +\left\lfloor \frac{\delta (k)}{\frac{1}{3^k-1}+\frac{1}{2^k-1}}\right\rfloor =2^k$ holds up to $k=350000.$
EDIT 29 Nov, 2017 New reductions.
Let exp3 = k $\geq2 \land (3^k - a)/(2^k) == (3^k - b - 1)/(2^k - 1) == c$, then we reduce using {a, b, c}, {b,c,a}, and {c,a,b} and back substitution. This results in three 3-variable diophantine equations to be proved: $$k>1\land b=\frac{a*2^k-a-2^k+3^k}{2^k}\land c=\frac{a \left(-2^k\right)+a-3^k+6^k}{2^k \left(2^k-1\right)}$$ $$k>1\land c=\frac{-b+3^k-1}{2^k-1}\land a=\frac{b*2^k+2^k-3^k}{2^k-1}$$ $$k>1\land a=\frac{c*2^k-c*2^{2 k}-3^k+6^k}{2^k-1}\land b=c \left(-2^k\right)+c+3^k-1$$

Answer 3

We look at the small difference $\delta(k)$ of the the two fractions (which become very near by higher $k$) $$\frac{3^k-1}{2^k-1}-\left(\frac{3}{2}\right)^k=\delta(k). \tag 1$$ Actually $\delta(k)$ is smaller than $1$ and approaches zero with higher $k$ which can be seen when expanded: $$\delta(k)=\frac{3^k-2^k}{2^k(2^k-1)} \lt 1 $$ and is of order$ (3/4)^k$.

If we find now, that also the fractional values of the terms in eq (1) equal $\delta(k)$ thus if we have that $$\left\{ \frac{3^k-1}{2^k-1} \right\}-\left\{ \left(\frac{3}{2}\right)^{k} \right\}= \delta(k), \tag 2$$ then it is obvious, that the two terms in (1) have also a common floor and we can write $$\left\lfloor {3^k-1 \over 2^k-1}\right\rfloor=\left\lfloor\left({3 \over 2}\right)^k\right\rfloor. \tag 3$$

The truth of this equality (3) for all $k \geq 2$ is what we want to show.

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We introduce now shorter notations $$D(k)=\delta(k) \cdot 2^k =\frac{3^k-2^k}{2^k-1}=\frac{3^k-1}{2^k-1} -1 \tag {4.1}$$ and $$ E(k)=\delta(k) \cdot \frac{(3^k-1)(2^k-1)}{3^k-2^k} = {3^k-1\over 2^k} \tag {4.2} $$ and conjecture, that there is always an integer $c$ between them: $$D(k) \lt c \le E(k) \tag 5$$

If that is indeed the case then we can write $$\therefore\ c= \left\lfloor\frac{3^k-1}{2^k-1}\right\rfloor=\left\lfloor\left(\frac{3}{2}\right)^k\right\rfloor \text{for }k\geq2.\ \square$$

(Note: we didn't the required proof here, which actually seems out of reach. The motivation of this "answer" was just to make the original OP's claim and its ideas nicer to read)