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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 $k=350000.$

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: $$k>1\land b=\frac{a*2^k-a-2^k+3^k}{2^k}\land c=\frac{a \left(-2^k\right)+a-3^k+6^k}{2^k \left(2^k-1\right)}$$ $$k>1\land c=\frac{-b+3^k-1}{2^k-1}\land a=\frac{b*2^k+2^k-3^k}{2^k-1}$$ $$k>1\land a=\frac{c*2^k-c*2^{2 k}-3^k+6^k}{2^k-1}\land b=c \left(-2^k\right)+c+3^k-1$$

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,
$g(k)=2^k+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor-2\ \ \ \ \text{ if }2^k \left\lbrace\left( \frac{3}{2} \right)^k \right\rbrace+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor\leq 2$,
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. Next, we substitute the upper boundaries for $a$ and $c$, and add a small fudge factor, $(1-\delta (k))$ to make it provable: $ 2^k (1-\delta (k)) + \frac{\delta (k)}{\frac{1}{3^k-1}+\frac{1}{2^k-1}} <2^k+(1-\delta (k))\iff k\geq1\land k \in \mathbb{Z}.$
Then we substitute the lower boundary for $c$, remove the fudge factor, 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 trivally true, which affirms that the boundaries are rigid.

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 $k=350000.$

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 $k=350000.$

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,
$g(k)=2^k+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor-2\ \ \ \ \text{ if }2^k \left\lbrace\left( \frac{3}{2} \right)^k \right\rbrace+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor\leq 2$,
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. Next, we substitute the upper boundaries for $a$ and $c$, and add a small fudge factor to make it provable: $ 2^k (1-\delta (k)) + \frac{\delta (k)}{\frac{1}{3^k-1}+\frac{1}{2^k-1}} <2^k+1\iff k\geq1\land k \in \mathbb{Z}.$
Empirically, we find that the fudge factor cannot be less than 1. Then we substitute the lower boundary for $c$, remove the fudge factor, 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: $\left\lfloor 2^k (1-\delta (k))\right\rfloor +\left\lfloor \frac{\delta (k)}{\frac{1}{3^k-1}+\frac{1}{2^k-1}}\right\rfloor =2^k$ holds up to $k=350000.$

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,
$g(k)=2^k+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor-2\ \ \ \ \text{ if }2^k \left\lbrace\left( \frac{3}{2} \right)^k \right\rbrace+\left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor\leq 2$,
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. Next, we substitute the upper boundaries for $a$ and $c$, and add a small fudge factor, $(1-\delta (k))$ to make it provable: $ 2^k (1-\delta (k)) + \frac{\delta (k)}{\frac{1}{3^k-1}+\frac{1}{2^k-1}} <2^k+(1-\delta (k))\iff k\geq1\land k \in \mathbb{Z}.$
Then we substitute the lower boundary for $c$, remove the fudge factor, 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 trivally true, which affirms that the boundaries are rigid.

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 $k=350000.$