I'm looking to find the finite set of integers that satisfy this equation:
$$\frac{3^{a-1}+2c}{3^{a}-2^{b}}\in\mathbb{Z} \text{ where a,b,c}\in \mathbb{Z}\text{ , }b\geq a\gt 0$$
Value a represents the cardinality of sequence P where $P_n\in\mathbb{N}^+$, while value b represents $\sum P$
Sequence $(Q_n) =\sum _{i=1}^n P_i$. bounds of 2c are defined as
$$2c=\sum_{j=0}^{\#P-1}3^j2^{\sum _{k=1}^{\#P-j-1}P_k} $$
Expanded power of 2 term for readability
$$2^x \text{ where } x= \sum _{k=1}^{\#P-j-1}P_k$$
Brute force has found 11 values.
Looking for a help to create a generalized form to prove this set of 11 values is the exhaustive list in the domain of integers.
Examples that are known:
(a=1,b=1,c=0) = 1 P=(1)
(a=1,b=2,c=0) = -1 P=(2)
(a=2,b=3,c=1) = 5 P=(1,2)
(a=2,b=3,c=2) = 7 P=(2,1)
(a=7,b=11,c=817) = 17 P=(1,1,1,2,1,1,4)
(a=7,b=11,c=1373) = 25 P=(1,1,2,1,1,4,1)
(a=7,b=11,c=2207) = 37 P=(1,2,1,1,4,1,1)
(a=7,b=11,c=2485) = 41 P=(2,1,1,4,1,1,1)
(a=7,b=11,c=3458) = 55 P=(1,1,4,1,1,1,2)
(a=7,b=11,c=3874) = 61 P=(1,4,1,1,1,2,1)
(a=7,b=11,c=5960) = 91 P=(4,1,1,1,2,1,1)
I have already proved that for a=1 the given 2 values are exhaustive since when a=1 and b>2 and the restriction on c $$\frac{3^0+2c}{3^1-2^{b>2}}=\frac{1+2c}{3-2^{3}}\notin\mathbb{Z}$$
EDIT: Additional Clarification: Sequence P and limits on c
I appreciate the effort of the solution by N.S. It works great, except for the constraint on possible valid c values. This happened because I wasn't clear as I could have been on what C represents, and did a poor job of detailing the limits on c, including why the called out "missed" example of a=1,b=1,k=3 is not valid with the additional constraint.
The formula above represents a simplification of the problem I am working on, which involves finite sequence P where $P_n =\mathbb{Z}^+$ and values a, b, and ultimately c are derived from this sequence.
a=#P=cardinality of P
b=$\sum P$
For clarity, let series Q be defined as the partial sums of P such that $$Q_n=\sum _{i=1}^{n}P_i$$ and define function R(n) such that $$ R(n)=\left\{\begin{matrix} Q_{\#Q-n-1} & n<\#Q-1 \\ 0 & n\geq\#Q-1 \end{matrix}\right. $$ So function R(n) returns Q elements 1 to #Q-1 in reverse order, and returns $0$ for $Q_{\#Q}$
The numerator of the function $3^{a-1}+2c$ as above was not clearly stated to be a simplification of the following sum:
$$
\sum _{i=0}^{\#P=1}3^i2^{R(i)}
$$
For example, a sequence P with cardinality of 7 expands to
$$
3^02^{Q_6} + 3^12^{Q_5} + 3^22^{Q_4} + 3^32^{Q_3} + 3^42^{Q_2} + 3^52^{Q_1} + 3^62^{0}
$$
Which I (over?) simplified by grouping all the $Q_n$ terms as $2c$ separate from $3^62^0$, resulting in the numerator of the original equation. In the particular case where P is only 1 element, the expansion doesn't have terms for 2c, so c must be 0.
This expansion does reveal a bounds and actual finite set of possible values for $2c$ for any selected a and b. Continuing from the example above, $a=\#P=7$ and given $b=Q_7=11$ means there is a maximum and minimum bound value for 2c.
Max (a=7,b=11)
P = (5,1,1,1,1,1,1)
Q = (5,6,7,8,9,10,11)
2c = 21280, c=10640
Min (a=7,b=11)
P = (1,1,1,1,1,1,5)
Q = (1,2,3,4,5,6,11)
2c = 1330, c = 665
There are 5 possible unique combinations of positive integers that will sum to 11, and a total of 210 unique combinations of those integers that result in unique c values.
5 combinations for P and combinations
(5,1,1,1,1,1,1) : 7
(4,2,1,1,1,1,1) : 42
(3,3,1,1,1,1,1) : 21
(3,2,2,1,1,1,1) : 105
(2,2,2,2,1,1,1) : 35
Total combinations 210
Unless necessary, I won't enumerate the 210 different valid c value....
I hope this additional clarification brings more insight into the problem I am struggling with.
The way you set the problem, there are infinitely many such triples.
Indeed, fix an arbitrary odd $k \in \mathbb Z$. Pick $b \geq a > 0$ and set $$c=\frac{3k-1}{2}3^{a-1}-k2^{b-1}$$
Then $a,b,c$ is a solution.
Note that it is easy to argue that $(a,b, \frac{3k-1}{2}3^{a-1}-k2^{b-1})$ describe all such solutions.
Note that $$ \frac{3^{a-1}+2c}{3^{a}-2^{b}} =\frac{3^{a-1}+((3k-1)3^{a-1}+2k2^{b-1}}{3^{a}-2^{b}}\\ =\frac{3^{a-1}+(3k)3^{a-1} -3^{a-1}+2k2^{b-1}}{3^{a}-2^{b}}=\frac{(3k)3^{a-1} +k2^{b}}{3^{a}-2^{b}}=\frac{k3^{a} +k2^{b}}{3^{a}-2^{b}}=k $$
P.S. Note that $a=1,b=1, k=1$ leads to $c=0$ as in your first solution.
$a=1,b=2, k=-1$ leads to $c=0$.
But, $a=1,b=1, k=3$ leads to $c=1$, which is a solution you missed.
P.P.S. Setting $k=2m+1$ your solution becomes $$c=(3m+1)3^{a-1}-(2m+1)2^{b-1}=m \cdot 3^a +3^{a-1}-m \cdot 2^b -2^{b-1}$$ which is probably easier to use. This works for all $m \in \mathbb Z$.