Explanation:
Okay, so I need to compute the sum over all triples (a, b, c) of positive integers that can form a triangle with sides a, b, c, of the term 2^a / (3^b * 5^c). Hmm. Let me think about how to approach this.

First, let's recall that for three positive integers a, b, c to form a triangle, they must satisfy the triangle inequalities. That is, the sum of any two sides must be greater than the third side. So, for a, b, c, we need:

a + b > c,
a + c > b,
b + c > a.

But since a, b, c are positive integers, maybe we can find a way to express these inequalities in terms of each variable being less than the sum of the other two. Alternatively, maybe we can fix an order to some extent. For example, without loss of generality, assume that a ≤ b ≤ c. Then, the triangle inequality reduces to a + b > c, since the other inequalities would automatically hold. For example, if a ≤ b ≤ c, then a + c > b is always true because c ≥ b and a is positive, so a + c ≥ a + b ≥ b + 1 > b. Similarly, b + c > a is also always true because b and c are both at least a, so their sum is at least 2a ≥ a. So the only non-trivial condition is a + b > c when a ≤ b ≤ c.

Therefore, maybe I can compute the sum by considering all triples (a, b, c) with a ≤ b ≤ c and a + b > c, then multiply by some symmetry factor? Wait, but the original problem doesn't specify any ordering, so each triple (a, b, c) can be in any order. However, since the variables a, b, c are in the exponents of different bases (2^a, 3^b, 5^c), the terms aren't symmetric. So permutations of (a, b, c) would lead to different terms. Therefore, maybe it's better to think of the problem as considering all triples where the triangle inequalities hold, regardless of order.

But that seems complicated. Maybe there's a generating function approach or inclusion-exclusion?

Alternatively, maybe I can think of the sum over all a, b, c ≥ 1 of 2^a / (3^b 5^c) where the triangle inequalities hold. So the total sum is equal to the sum over all triples (a, b, c) with a, b, c ≥ 1, of 2^a / (3^b 5^c) multiplied by an indicator function that is 1 if a, b, c can form a triangle, and 0 otherwise.

Therefore, the problem reduces to computing:

S = ∑_{a=1}^∞ ∑_{b=1}^∞ ∑_{c=1}^∞ [indicator(a, b, c satisfy triangle inequalities)] * (2^a / (3^b 5^c))

But how to compute this sum? It might be helpful to use the triangle inequalities and express the indicator as constraints on a, b, c. Since triangle inequalities are symmetric, maybe we can exploit some symmetry.

Alternatively, maybe we can compute the total sum over all a, b, c and subtract the ones that don't satisfy the triangle inequalities. That is:

Total sum without any restrictions is S_total = (∑_{a=1}^∞ 2^a) * (∑_{b=1}^∞ 1/3^b) * (∑_{c=1}^∞ 1/5^c).

But wait, but the original problem is not the total sum, but only the sum over triples that satisfy the triangle inequalities. So maybe:

S = S_total - S_bad, where S_bad is the sum over all triples that don't satisfy the triangle inequalities.

But how do I compute S_bad? Well, the triangle inequalities are not satisfied if at least one of the following is true:

a + b ≤ c,
a + c ≤ b,
b + c ≤ a.

So, S_bad is the sum over triples where at least one of these inequalities holds. But inclusion-exclusion might be needed here.

Alternatively, maybe we can compute S as follows:

First, compute S_total, then subtract the cases where the triangle inequalities are not satisfied. However, inclusion-exclusion is necessary because some triples might be counted multiple times if they fail more than one inequality.

So let me write that down.

Let me define:

S_total = (∑_{a=1}^∞ 2^a) * (∑_{b=1}^∞ 1/3^b) * (∑_{c=1}^∞ 1/5^c)

Compute each sum:

Sum_{a=1}^∞ 2^a = 2/(1 - 2) - 1? Wait, geometric series formula is sum_{n=0}^∞ r^n = 1/(1 - r). So sum from n=1 to ∞ is 1/(1 - r) - 1 = r/(1 - r). Therefore:

Sum_{a=1}^∞ 2^a = 2/(1 - 2) = -2? Wait, that's impossible, since sum_{a=1}^∞ 2^a diverges. Wait, but the original problem says that (a, b, c) are positive integers, so the term 2^a / (3^b 5^c) would have 2^a growing exponentially, but 3^b and 5^c are also exponential. However, the sum over all a, b, c might still diverge? Wait, but the problem says to compute the sum over all such triples. However, if the total sum diverges, then we have to be careful. However, since we need to compute only the sum over triples that satisfy the triangle inequalities, maybe the convergence is better?

Wait, let me check the convergence. Let me think about the total sum S_total:

Sum_{a=1}^∞ 2^a * Sum_{b=1}^∞ 1/3^b * Sum_{c=1}^∞ 1/5^c.

Each of these sums individually diverges? Wait, Sum_{a=1}^∞ 2^a is divergent. But in reality, the original problem must have a convergent sum because otherwise the problem would be ill-posed. However, since the term is 2^a / (3^b 5^c), which has different exponents for a, b, c. Wait, if we think of each term as 2^a * 3^{-b} * 5^{-c}, then the total sum is a product of three independent sums? Wait, no, the triangle inequalities impose some relation between a, b, c. However, without any constraints, the total sum is divergent because the 2^a term is divergent. Therefore, perhaps my approach is wrong?

But the problem states to compute the sum over all triples (a,b,c) that can form a triangle. However, if there are infinitely many such triples, and the terms don't go to zero fast enough, then the sum may diverge. But maybe due to the triangle condition, the sum converges? Hmm.

Wait, let me check for convergence. Suppose we fix a, b, c. The term is 2^a / (3^b 5^c). If we fix a and b, then as c increases, the term decreases as 1/5^c. However, for triangle inequalities, when we consider that for fixed a and b, c must be less than a + b. So for fixed a and b, the sum over c is sum_{c=1}^{a + b - 1} 1/5^c. Which converges as a finite sum. Similarly, if we fix other variables. Therefore, perhaps the total sum converges.

Wait, maybe I need to re-express the problem with the triangle inequalities.

But earlier, I thought about S_total, but maybe instead of considering all triples, I need to consider the sum with the triangle inequalities. Let me think again.

Let me think of the problem as first considering all a, b, c ≥ 1, and then subtracting those which don't satisfy triangle inequalities. However, since the total sum diverges (as sum_{a} 2^a is divergent), perhaps I need a different approach. Maybe instead of considering the entire sum, I can find a generating function that encodes the triangle conditions.

Alternatively, maybe I can fix an ordering and use symmetry. Earlier, I thought that if I assume a ≤ b ≤ c, then the triangle inequality reduces to a + b > c. Then, perhaps the original sum can be written as 6 times the sum over all a ≤ b ≤ c with a + b > c of the term 2^a / (3^b 5^c), plus some other terms? Wait, no, because the terms are not symmetric. Since the exponents are different for a, b, c, the permutations would give different terms. Therefore, we cannot just multiply by symmetry factors. For example, if a ≤ b ≤ c, then a, b, c could be permuted in different ways, but each permutation would have different values of 2^a / (3^b 5^c). Therefore, perhaps this approach is not straightforward.

Alternatively, let me try to consider the triangle inequalities in terms of variables. Let me think of the three triangle inequalities:

1. a + b > c
2. a + c > b
3. b + c > a

But since a, b, c are positive integers, we can think of the conditions as:

For all permutations, the sum of the two smaller sides must exceed the largest side.

Therefore, if I can find a way to count all triples (a, b, c) where the largest side is less than the sum of the other two. Therefore, perhaps if I consider for each triple, find which one is the largest. But again, the problem is the asymmetry in exponents.

Alternatively, maybe I can use generating functions with constraints.

Let me consider generating functions where I can encode the triangle inequalities.

But I need to think of variables a, b, c. Let me think about how to model the triangle inequalities.

Alternatively, perhaps I can fix the maximum value among a, b, c. Let's assume that c is the maximum. So c ≥ a, c ≥ b. Then, the triangle inequality requires that a + b > c. Similarly, if a is the maximum, then b + c > a, and if b is the maximum, then a + c > b. However, since the exponents are different, maybe it's better to consider each case where a particular variable is the maximum.

But again, since the exponents differ, this complicates things. Let me try.

Case 1: c is the largest side. So c ≥ a, c ≥ b, and a + b > c.

Case 2: b is the largest side. So b ≥ a, b ≥ c, and a + c > b.

Case 3: a is the largest side. So a ≥ b, a ≥ c, and b + c > a.

But since the triples (a, b, c) can be in any order, and the exponents differ, these cases will contribute different amounts to the sum. However, since the problem is symmetric with respect to permutations of a, b, c only if the exponents were the same. But here, since 2^a / (3^b 5^c) is not symmetric, we can't just multiply by the number of permutations. Therefore, perhaps we need to handle each case separately.

Alternatively, maybe we can think of the problem as follows:

For each triple (a, b, c) where the triangle inequalities hold, the term is 2^a / (3^b 5^c). So, if I can find a way to express this sum by considering for each possible maximum side, and then sum over all such possibilities.

Alternatively, maybe it's better to express the sum as the product of generating functions with constraints.

But I need to think of another way. Let me recall that the original problem is similar to counting the number of triangles with given side lengths, but with weights. So perhaps I need to find a way to express the conditions a + b > c, etc., in terms of generating functions.

Alternatively, think about the sum over all a, b, c ≥ 1 with a + b > c, a + c > b, and b + c > a. Let me try to express this as:

Sum_{a, b, c ≥ 1} [indicator(a + b > c and a + c > b and b + c > a)] * (2^a / (3^b 5^c)).

Which can be written as:

Sum_{a, b, c ≥ 1} (2^a / (3^b 5^c)) * [indicator(a + b > c) * indicator(a + c > b) * indicator(b + c > a)].

But how to compute this?

Alternatively, note that the triangle inequalities are equivalent to:

max(a, b, c) < a + b + c - max(a, b, c).

Wait, because for any triangle, the sum of the two smaller sides must exceed the largest side. Let me check. Suppose c is the largest side, then a + b > c. Since c > a and c > b, then a + b > c. But then the sum of all sides is a + b + c, so a + b = (a + b + c) - c. Therefore, the condition is that the largest side is less than the sum of the other two sides, which is equivalent to the largest side being less than half of the total perimeter.

Wait, more precisely, since if we denote S = a + b + c, then the largest side is less than S - largest side, so 2 * largest side < S. Therefore, largest side < S / 2. Therefore, the triangle inequalities are equivalent to the largest side being less than half of the total perimeter. Therefore, perhaps we can use this.

Therefore, if we let M = max(a, b, c), then the condition is M < (a + b + c)/2.

Alternatively, 2M < a + b + c.

But maybe this is helpful? Not sure yet.

Alternatively, since the problem is multiplicative, maybe I can express the sum as follows:

Let me think about fixing M, the maximum of a, b, c, and then sum over all triples where max(a, b, c) = M and the triangle inequalities hold. But again, the problem is the asymmetry in the exponents.

Alternatively, maybe I can think of generating functions where I consider the variables a, b, c, with constraints.

Wait, here's an idea. Let me consider generating functions for a, b, c with constraints. Let me define generating functions for each variable:

For variable a: it can be any positive integer, but with some conditions depending on b and c.

But since the triangle inequalities involve mutual relations between a, b, c, maybe it's complicated.

Alternatively, since we can consider the problem in terms of variables with ordering. Suppose we first fix an ordering, say a ≤ b ≤ c, then the triangle condition is a + b > c. Then, the total sum over all triples (a, b, c) can be expressed as the sum over all such ordered triples multiplied by the number of permutations that correspond to their different orderings. However, again, since the exponents are different, this complicates things.

Wait, perhaps not. Let's suppose that we consider ordered triples where a ≤ b ≤ c and a + b > c. Then, each such triple corresponds to certain permutations. However, because of the different exponents, the contribution of each permutation is different. For example, if we have a triple (a, b, c) with a < b < c, then the permutations where a is in different positions would have different exponents.

Therefore, maybe it's better to think that each triple (a, b, c) which satisfies the triangle inequalities can be in any order, but since the exponents are different, we need to consider each case.

Alternatively, maybe think of the original problem as:

Let me consider the sum over all a, b, c where a + b > c, a + c > b, and b + c > a. Since these are symmetric conditions, perhaps we can use some sort of generating function where we account for all three inequalities.

Alternatively, note that the three inequalities can be combined as:

For all i, j, k, with i, j, k being permutations of a, b, c: i + j > k.

But I don't know if that helps.

Alternatively, since each inequality is a + b > c, etc., we can note that for all triples (a, b, c), the triangle inequalities are equivalent to:

c < a + b,

b < a + c,

a < b + c.

But if we can express these as constraints on the variables. For example, if we fix c, then for a and b, we need a + b > c, and also a and b must be such that they don't violate other inequalities. Wait, but since a, b, c are symmetric except for the exponents, maybe we can fix c and consider the sum over a, b such that a + b > c, and then also considering the other inequalities? Wait, but if c is fixed, then the other inequalities would be a < b + c and b < a + c. Which are automatically satisfied if a and b are positive integers. Because a < b + c is equivalent to a - b < c. Since a and b are positive integers, unless a is very large compared to b and c. Wait, but for example, if c is fixed, and suppose a is very large, say a ≥ b + c + 1, then a < b + c would not hold. Therefore, that would be a case where a is too large. Therefore, so if we fix c, then to satisfy all triangle inequalities, we need:

a + b > c,

a < b + c,

b < a + c.

But if c is the largest side, then the first inequality is the only one needed. However, if c is not the largest side, then another inequality may be necessary.

This seems too vague. Maybe we need another approach.

Let me think about the following idea. Suppose I first consider all triples (a, b, c) with a, b, c ≥ 1, and then subtract those that do not satisfy the triangle conditions. However, as I thought earlier, the total sum diverges. However, perhaps the divergent parts cancel out?

Wait, but if the total sum diverges and the bad sum also diverges, then their difference might converge?

Alternatively, maybe there's another way. Let me think about the sum over all triples (a, b, c) with triangle inequalities. Since each triple must satisfy a + b > c, a + c > b, b + c > a. Let me try to find generating functions for each of these.

Alternatively, notice that the product of the three inequalities can be represented as follows:

For each triple (a, b, c), the triangle inequalities are equivalent to:

a + b + c > 2 * max(a, b, c).

Let me check this. Suppose that the maximum of a, b, c is M. Then the triangle inequalities are equivalent to M < a + b + c - M, which simplifies to 2M < a + b + c.

Therefore, the triangle inequalities are equivalent to 2 * max(a, b, c) < a + b + c. Therefore, perhaps we can use this condition to compute the sum.

Therefore, we can write the desired sum as:

Sum_{a, b, c ≥ 1} [indicator(2 * max(a, b, c) < a + b + c)] * (2^a / (3^b 5^c)).

Now, perhaps I can express this sum as follows: Let me consider cases depending on which variable is the maximum. Let me split the sum into three cases:

Case 1: a is the maximum, i.e., a ≥ b, a ≥ c.

Case 2: b is the maximum, i.e., b ≥ a, b ≥ c.

Case 3: c is the maximum, i.e., c ≥ a, c ≥ b.

Note that these cases are mutually exclusive and cover all possibilities where exactly one variable is the maximum. However, there are also cases where two variables are equal and the third is less, or all three are equal. But since the maximum can be tied (for example, a = b ≥ c), but in that case, we still have a single maximum (or two maxima). However, since we need to consider all triples where the maximum is less than the sum of the other two sides. Therefore, perhaps we can compute each case separately.

However, since the exponents are different for a, b, c, the contribution from each case is different. Therefore, I need to compute each case:

Let me define:

S1 = sum over a ≥ b, a ≥ c, and 2a < a + b + c of 2^a / (3^b 5^c)

S2 = sum over b ≥ a, b ≥ c, and 2b < a + b + c of 2^a / (3^b 5^c)

S3 = sum over c ≥ a, c ≥ b, and 2c < a + b + c of 2^a / (3^b 5^c)

Then, the total sum S = S1 + S2 + S3. However, we need to be careful not to double count cases where two variables are equal and the third is equal or less. But because of the max condition, each triple is counted exactly once in one of the cases. For example, if a = b = c, then all three cases would include this triple? Wait, no. For example, if all three are equal, then all three cases would satisfy the conditions. But actually, when considering case 1, a is the maximum, but if a = b = c, then a is equal to the other variables. However, in the definition of Case 1, it's "a ≥ b, a ≥ c". So if a = b = c, then the conditions are satisfied. But in this case, does it get counted in all three cases? No, because for each case, the maximum is considered. However, in reality, if all three variables are equal, then all three cases would include it? Wait, no. If we define S1 as a ≥ b, a ≥ c, then the triple (a, a, a) would be included in S1, S2, S3? Wait, no. For example, S1 is for when a is the maximum. If a = b = c, then a is the maximum. But also, b is the maximum and c is the maximum. So in reality, this triple would be counted in all three S1, S2, S3. Therefore, we would be overcounting. Therefore, this approach may not work unless we adjust for overlapping cases.

Therefore, to avoid overcounting, perhaps we should consider the cases where a, b, c are distinct, and then handle the equality cases separately. However, this complicates things further. Alternatively, perhaps it's better to use inclusion-exclusion, but I need to think.

Alternatively, maybe instead of splitting into cases where a, b, or c is the maximum, we can use the following approach. Let me consider that for each triple (a, b, c), the maximum of a, b, c is M. Then, the sum can be written as:

S = ∑_{M=1}^∞ ∑_{a,b,c with max(a,b,c)=M and 2M < a + b + c} (2^a / (3^b 5^c))

But even this seems complicated. However, perhaps for a given M, we can compute the sum over all triples (a, b, c) where max(a, b, c) = M and 2M < a + b + c.

But for each M, the condition max(a, b, c) = M implies that at least one of a, b, c is equal to M, and the others are ≤ M.

Furthermore, the triangle condition is 2M < a + b + c. Since max(a, b, c) = M, the sum a + b + c is at least M + 1 + 1 = M + 2 (if two other sides are 1) and at most 3M (if all sides are M). Therefore, the triangle condition 2M < a + b + c implies that a + b + c > 2M. Since a + b + c ≤ 3M, this condition becomes 2M < a + b + c ≤ 3M. Therefore, for each M, we have to sum over all triples (a, b, c) with max(a, b, c) = M and a + b + c > 2M.

But how can we compute this?

Let me fix M. Let me consider the sum over all (a, b, c) with max(a, b, c) = M and a + b + c > 2M.

First, let's compute the total number of triples with max(a, b, c) = M. Then, subtract those which have a + b + c ≤ 2M. However, since our problem is not about counting but about summing weighted terms, let me think.

But the problem is that the sum is over (a, b, c) with max(a, b, c) = M and a + b + c > 2M. So for each M, we need to compute:

Sum_{a, b, c, max(a,b,c)=M} [indicator(a + b + c > 2M)] * (2^a / (3^b 5^c))

But this seems challenging. However, perhaps I can find this sum by considering that for each M, the triples with max(a, b, c) = M can be partitioned into different cases depending on which variable is equal to M. Let me try:

Case 1: a = M, and b ≤ M, c ≤ M, and neither b nor c is equal to M. Then, the condition is max(a, b, c) = M.

Case 2: b = M, and a ≤ M, c ≤ M, and neither a nor c is equal to M.

Case 3: c = M, and a ≤ M, b ≤ M, and neither a nor b is equal to M.

Case 4: a = M, b = M, c < M.

Case 5: a = M, c = M, b < M.

Case 6: b = M, c = M, a < M.

Case 7: a = M, b = M, c = M.

But this is getting complicated, but maybe manageable.

However, since the exponents are different, perhaps we need to handle each case with its own contribution.

Alternatively, maybe we can think of all triples with max(a, b, c) = M. Let me note that the sum over all triples with max(a, b, c) = M is equal to:

Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M [indicator(max(a,b,c)=M)] * (2^a / (3^b 5^c))

But this is equal to:

Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M (2^a / (3^b 5^c)) * [indicator(max(a,b,c)=M)]

This can be rewritten as:

Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M (2^a / (3^b 5^c)) * [indicator(a = M or b = M or c = M) - indicator(a = M and b = M) - ... + ... ]

But inclusion-exclusion here is going to get messy. Alternatively, note that the number of triples where max(a,b,c)=M is equal to 3*(M^2) - 3*M + 1, but again with weights.

Alternatively, maybe we can use the principle that the sum over max(a,b,c)=M is equal to the sum over all triples with a, b, c ≤ M, and subtract those with max(a,b,c) < M. Therefore:

Sum_{max(a,b,c)=M} f(a,b,c) = Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M f(a,b,c) - Sum_{max(a,b,c) < M} f(a,b,c)

But the second term is Sum_{a=1}^{M-1} Sum_{b=1}^{M-1} Sum_{c=1}^{M-1} f(a,b,c)

Therefore, for our problem, the sum over all triples with max(a,b,c)=M and a + b + c > 2M is equal to:

[Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M f(a,b,c) * indicator(max(a,b,c)=M)] - [Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M f(a,b,c) * indicator(max(a,b,c)=M) * indicator(a + b + c ≤ 2M)]

But maybe it's better to compute it as:

For each M, the sum over triples with max(a,b,c) = M is:

Total = [sum_{a=1}^M 2^a] * [sum_{b=1}^M 1/3^b] * [sum_{c=1}^M 1/5^c] - [sum_{a=1}^{M-1} 2^a] * [sum_{b=1}^{M-1} 1/3^b] * [sum_{c=1}^{M-1} 1/5^c]

But this seems complex, but maybe manageable. However, we need to then subtract from this the sum over those triples where a + b + c ≤ 2M. Therefore, it's getting very involved.

Alternatively, maybe it's better to think of the original problem as follows. Since the triangle conditions are symmetric, perhaps I can use generating functions with constraints. Let me try.

Let me think of generating functions where I have variables a, b, c with a + b > c, a + c > b, b + c > a. Let me note that these inequalities can be rewritten as:

c < a + b,

b < a + c,

a < b + c.

But in generating functions, how do I encode these inequalities? Maybe consider generating functions for each variable, but with constraints.

Alternatively, for each variable, we can think of generating functions and use the constraints. Let me consider the generating function for a, b, c.

But I need to think of generating functions that satisfy a + b > c, etc. This seems challenging. However, note that these inequalities can be combined into the single condition that the sum of the two smaller sides is greater than the largest side, as I thought earlier.

Therefore, let me consider that the generating function for each triple is 2^a / (3^b 5^c) multiplied by the indicator that the sum of the two smaller sides is greater than the largest side.

But how to model this?

Alternatively, here's another idea. Since the triangle inequalities are equivalent to the condition that no one side is greater than or equal to the sum of the other two. Therefore, the complement of this is that at least one side is greater than or equal to the sum of the other two. Therefore, we can write the sum S as:

S = Total - S_bad,

where Total is the sum over all triples (a, b, c) of 2^a / (3^b 5^c), and S_bad is the sum over triples where at least one side is ≥ the sum of the other two. But as before, Total is divergent. So this might not help, but maybe the divergences cancel out?

Alternatively, maybe I can compute S_total and S_bad with some regularization? But that seems too vague.

Alternatively, think of the problem as follows: For variables a, b, c, the triangle inequalities can be written as:

c < a + b,

b < a + c,

a < b + c.

Which is equivalent to:

For all permutations, the side is less than the sum of the other two.

Alternatively, notice that if we consider the three inequalities:

c < a + b,

a < b + c,

b < a + c.

These are symmetric. Therefore, maybe I can consider the generating function where I have the product of generating functions for each variable, multiplied by constraints.

Alternatively, consider that the three inequalities can be rewritten as:

c < a + b => c ≤ a + b - 1,

Similarly for others. Therefore, if I can write the generating function as:

Sum_{a,b,c ≥ 1} 2^a / (3^b 5^c) * [indicator(c ≤ a + b - 1) * indicator(a ≤ b + c - 1) * indicator(b ≤ a + c - 1)]

But how to compute this?

Alternatively, note that the three inequalities can be combined into a single condition. Let me think about fixing a and b, then c must satisfy both c < a + b and c > |a - b|. Wait, because for the triangle inequality, if we fix a and b, then c must satisfy |a - b| < c < a + b. But since c is a positive integer, c must satisfy max(1, |a - b| + 1) ≤ c ≤ a + b - 1.

Wait, but this is only if we fix a and b. However, the other inequalities also need to be considered. Wait, if we fix a and b, then c needs to satisfy c < a + b, and also, since a and b are fixed, the other inequalities:

a < b + c => c > a - b.

But since c is positive, this implies c ≥ max(1, a - b + 1).

Similarly, b < a + c => c > b - a => c ≥ max(1, b - a + 1).

Therefore, combining all, for fixed a and b, c must satisfy:

max(1, |a - b| + 1) ≤ c ≤ a + b - 1.

But since c ≥ 1, and |a - b| + 1 is at least 1 (since |a - b| is non-negative). Therefore, for fixed a and b, the range of c is from max(1, |a - b| + 1) to a + b - 1.

However, note that if a and b are such that a + b - 1 ≥ 1, which is true for a, b ≥ 1, except when a = b = 1, in which case c must be between max(1, 0 + 1) = 1 and 1 + 1 - 1 = 1. So c = 1. Which is correct, because for a = b = 1, c must be < 2, so c = 1. Which is valid.

Therefore, this seems promising. Therefore, the original sum can be rewritten as:

S = ∑_{a=1}^∞ ∑_{b=1}^∞ [∑_{c = max(1, |a - b| + 1)}^{a + b - 1} 2^a / (3^b 5^c)]

Similarly, since for each a and b, c is determined to be in that range. Therefore, the problem reduces to computing this double sum over a and b, and for each, summing over c. So maybe we can compute this.

So, let me focus on this. Let me denote:

For each a and b, compute the inner sum over c. Let me denote:

For fixed a and b, the inner sum is:

Sum_{c = L}^{U} 1/5^c, where L = max(1, |a - b| + 1), and U = a + b - 1.

Therefore, the total sum S is equal to:

Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * [Sum_{c = L}^{U} 1/5^c]

Therefore, if I can compute this inner sum over c, then multiply by 2^a / 3^b, and then sum over a and b, I can get the desired result.

Okay, let's try to compute this inner sum. Let me first handle the inner sum.

First, note that Sum_{c = L}^{U} 1/5^c is a geometric series from c = L to c = U. The formula for this is (1/5^L - 1/5^{U + 1}) / (1 - 1/5)) = (1/5^L - 1/5^{U + 1}) / (4/5) ) = 5/4 * (1/5^L - 1/5^{U + 1}).

Therefore, the inner sum is 5/4 * (1/5^L - 1/5^{U + 1}).

Therefore, substituting back, S becomes:

S = (5/4) * Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * [1/5^L - 1/5^{U + 1}]

Where L = max(1, |a - b| + 1) and U = a + b - 1.

So, S = (5/4) * [Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * (1/5^L - 1/5^{U + 1}) ]

Therefore, let me split this into two sums:

S = (5/4)[ S1 - S2 ], where

S1 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^L,

and

S2 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{U + 1}.

So, let me compute S1 and S2 separately.

First, let's compute S1 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{L}, where L = max(1, |a - b| + 1).

Similarly, S2 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{U + 1}, where U = a + b - 1, so U + 1 = a + b.

Therefore, S2 becomes Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{a + b} = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 5^a) * (1/3^b / 5^b) ) = Sum_{a=1}^∞ (2/5)^a * Sum_{b=1}^∞ (1/15)^b.

Compute this:

Sum_{a=1}^∞ (2/5)^a = (2/5) / (1 - 2/5) = (2/5) / (3/5) = 2/3.

Sum_{b=1}^∞ (1/15)^b = (1/15) / (1 - 1/15) = (1/15) / (14/15) = 1/14.

Therefore, S2 = (2/3) * (1/14) = 2/(42) = 1/21.

Okay, so S2 is 1/21. Great, that was straightforward.

Now, S1 is more complicated: Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{max(1, |a - b| + 1)}.

Let me think about how to handle this. The term max(1, |a - b| + 1) can be simplified. Let me note that |a - b| + 1 is always at least 1, since |a - b| ≥ 0. Therefore, max(1, |a - b| + 1) is just |a - b| + 1. Wait, no! Wait: if |a - b| + 1 ≥ 1, which it always is, because |a - b| ≥ 0, so |a - b| + 1 ≥ 1. Therefore, actually, L = max(1, |a - b| + 1) = |a - b| + 1. Wait, that seems contradictory to earlier reasoning. Wait, no, let me check. The earlier idea was that for fixed a and b, c must be ≥ max(1, |a - b| + 1). However, |a - b| + 1 is always ≥ 1 since |a - b| ≥ 0. Therefore, max(1, |a - b| + 1) is just |a - b| + 1. Therefore, the inner sum over c becomes from c = |a - b| + 1 to c = a + b - 1. Wait, but if |a - b| + 1 is equal to L, then why was there a max(1, ...)? Because if |a - b| + 1 could be less than 1, but since |a - b| ≥ 0, then |a - b| + 1 ≥ 1. Therefore, L = |a - b| + 1.

Therefore, S1 is Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1 / 5^{ |a - b| + 1 }.

So S1 = (1/5) * Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) / 5^{|a - b|}.

Simplify: S1 = (1/5) * Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 5^{-|a - b|}.

So, this seems like a double sum over a and b of (2^a / 3^b) * 5^{-|a - b|}. Let me try to compute this.

To compute this double sum, perhaps I can split it into cases where a ≥ b and a < b. Since |a - b| is symmetric.

Let me denote for a ≥ b, |a - b| = a - b.

For a < b, |a - b| = b - a.

Therefore, the double sum becomes:

Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 5^{-|a - b|} = Sum_{a=1}^∞ Sum_{b=1}^a (2^a / 3^b) * 5^{-(a - b)} + Sum_{a=1}^∞ Sum_{b=a+1}^∞ (2^a / 3^b) * 5^{-(b - a)}.

Let me handle these two sums separately.

First, let's compute the first sum: Sum_{a=1}^∞ Sum_{b=1}^a (2^a / 3^b) * 5^{-(a - b)}.

Let me change variables. Let k = a - b. Since b ranges from 1 to a, k ranges from 0 to a - 1. So for each a, we can write:

Sum_{b=1}^a (2^a / 3^b) * 5^{-k} where k = a - b.

But k = a - b => b = a - k.

Therefore, substituting:

Sum_{k=0}^{a - 1} (2^a / 3^{a - k}) * 5^{-k} = 2^a * Sum_{k=0}^{a - 1} (1/3^{a - k}) * (1/5^k) = 2^a * (1/3^a) * Sum_{k=0}^{a - 1} (3/5)^k.

Wait, let me check:

Since 1 / 3^{a - k} = 3^k / 3^a, so:

(1/3^{a - k}) * (1/5^k) = 3^k / 3^a * 1 / 5^k = (3/5)^k / 3^a. Therefore:

Sum_{k=0}^{a - 1} (3/5)^k / 3^a = (1 / 3^a) * Sum_{k=0}^{a - 1} (3/5)^k.

Therefore, the entire first term becomes:

Sum_{a=1}^∞ [ 2^a / 3^a * Sum_{k=0}^{a - 1} (3/5)^k ].

Similarly, let's compute the second sum: Sum_{a=1}^∞ Sum_{b=a+1}^∞ (2^a / 3^b) * 5^{-(b - a)}.

Here, let me change variables again. Let m = b - a. Since b ≥ a + 1, m ≥ 1. So for each a, b = a + m, m ≥ 1.

Then:

Sum_{a=1}^∞ Sum_{m=1}^∞ (2^a / 3^{a + m}) * 5^{-m} = Sum_{a=1}^∞ 2^a / 3^a * Sum_{m=1}^∞ (1 / 3^m) * (1 / 5^m) ) = Sum_{a=1}^∞ (2/3)^a * Sum_{m=1}^∞ (1 / 15)^m.

Because 1 / 3^m * 1 / 5^m = 1 / 15^m.

Therefore, the second sum is:

Sum_{a=1}^∞ (2/3)^a * [ (1/15) / (1 - 1/15) ) ] = Sum_{a=1}^∞ (2/3)^a * [ 1 / 14 ] = (1 / 14) * [ (2/3) / (1 - 2/3) ) ] = (1 / 14) * [ (2/3) / (1/3) ) ] = (1 / 14) * 2 = 2 / 14 = 1 / 7.

Okay, so the second sum is 1/7. Now, going back to the first sum:

First sum is Sum_{a=1}^∞ [ (2/3)^a * Sum_{k=0}^{a - 1} (3/5)^k ].

Let me compute this. Let me note that Sum_{k=0}^{a - 1} (3/5)^k is a finite geometric series. The sum is [1 - (3/5)^a] / [1 - 3/5] = [1 - (3/5)^a] / (2/5) ) = (5/2)[1 - (3/5)^a].

Therefore, the first sum becomes:

Sum_{a=1}^∞ (2/3)^a * (5/2)[1 - (3/5)^a] = (5/2) Sum_{a=1}^∞ (2/3)^a - (5/2) Sum_{a=1}^∞ (2/3)^a * (3/5)^a.

Compute each part:

First term: (5/2) * Sum_{a=1}^∞ (2/3)^a = (5/2) * [ (2/3) / (1 - 2/3) ) ] = (5/2) * [ (2/3) / (1/3) ) ] = (5/2) * 2 = 5.

Second term: (5/2) * Sum_{a=1}^∞ (2/3 * 3/5)^a = (5/2) * Sum_{a=1}^∞ (2/5)^a = (5/2) * [ (2/5) / (1 - 2/5) ) ] = (5/2) * [ (2/5) / (3/5) ) ] = (5/2) * (2/3) = 5/3.

Therefore, the first sum is 5 - 5/3 = 10/3.

Therefore, combining both sums, the total double sum (for S1) is 10/3 + 1/7 = (70/21 + 3/21) = 73/21.

Wait, no: the first sum was 10/3 and the second sum was 1/7, so total double sum is 10/3 + 1/7 = (70 + 3)/21 = 73/21.

Therefore, S1 = (1/5) * 73/21 = 73 / 105.

Therefore, S = (5/4)(S1 - S2) = (5/4)(73/105 - 1/21).

Compute 73/105 - 1/21:

Convert 1/21 to 5/105, so 73/105 - 5/105 = 68/105.

Then, S = (5/4)(68/105) = (5 * 68) / (4 * 105) = (340) / 420 = Simplify numerator and denominator by 20: 17 / 21. Wait:

Wait, 5*68 = 340, 4*105 = 420.

Divide numerator and denominator by 20: 340 / 420 = 17 / 21. Yes.

Therefore, S = 17/21.

Wait, that seems too clean. Let me check the calculations again.

First, S1 was calculated as (1/5) * 73/21, which is 73/105. Then S2 is 1/21. So S1 - S2 is 73/105 - 5/105 = 68/105. Then multiply by 5/4: (68/105) * (5/4) = (68 * 5) / (105 * 4) = 340 / 420 = simplify by dividing numerator and denominator by 20: 17 / 21. Yes, that's correct.

So the final answer is 17/21?

Wait, but let me check again the steps for errors.

First, S2 was calculated as 1/21. Let me confirm:

For S2, we had:

S2 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1 / 5^{a + b}.

This is equal to [Sum_{a=1}^∞ (2/5)^a] * [Sum_{b=1}^∞ (1/15)^b].

Wait, hold on. Wait, 1/5^{a + b} = 1/5^a * 1/5^b. Therefore, the double sum becomes [Sum_{a} (2^a / 5^a)] * [Sum_b (1 / 3^b / 5^b)] ?

Wait, no. Wait, original expression:

(2^a / 3^b) * 1 / 5^{a + b} = (2^a / 5^a) * (1 / 3^b / 5^b) ?

Wait, no. It's (2^a / 3^b) * 1 / 5^{a + b} = 2^a / (3^b * 5^{a + b}) ) = (2 / 5)^a * (1 / (3 * 5)^b) ) = (2/5)^a * (1/15)^b. Therefore, the sum is [Sum_{a=1}^∞ (2/5)^a] * [Sum_{b=1}^∞ (1/15)^b]. Which is (2/3) * (1/14) = 1/21. Correct.

Then for S1, we had that the double sum became 73/21 and then multiplied by 1/5 gives 73/105. Then S1 - S2 = 73/105 - 5/105 = 68/105. Multiply by 5/4 gives 17/21. Seems correct.

But the problem says to compute the sum over all triples (a,b,c) that can form a triangle of 2^a / (3^b 5^c). So according to this calculation, the answer is 17/21. Is that possible?

But let me check if the logic is correct. When I broke the sum into S1 and S2, and then calculated them, the steps seem valid. But to be thorough, let me check with small values.

Alternatively, let me think about the total sum S_total. Earlier, we thought that it diverges, but in our calculation, we subtracted S2 from S1 and got a finite value. However, since both S1 and S2 were calculated as finite sums (due to the constraints on c), maybe it works out. Also, since we transformed the original problem into a convergent sum, it's possible.

Therefore, maybe the answer is 17/21. However, let me check with a simple case.

For example, consider all possible small triples (a, b, c) and compute the sum manually, then compare with the result.

But since the answer is a fraction, and we arrived at 17/21, let me see if there is any chance of error in the steps.

First, when calculating S1, we split into a ≥ b and a < b. Then converted the first sum into a double sum over a and k, etc. Then we found the first sum as 10/3 and the second sum as 1/7, totaling 73/21. Then S1 is that times 1/5: 73/105. Then S2 is 1/21. Then (S1 - S2) * 5/4 = 17/21. Seems correct.

Alternatively, perhaps I can check with M = 1, compute the contribution.

When M = 1, the only possible (a, b, c) with max(a, b, c) = 1 is (1,1,1). The triangle inequality condition is 1 + 1 > 1, which holds. Therefore, the contribution is 2^1 / (3^1 5^1) = 2/(3*5) = 2/15.

But according to our formula, let's check what part of the overall sum comes from M=1. Wait, but our derivation was based on a different approach, so maybe it's not straightforward.

Alternatively, since our final answer is 17/21 ≈ 0.8095, which is less than 1, which seems plausible.

Alternatively, let me think of the total sum over all (a,b,c) without triangle conditions. Wait, earlier we thought it was divergent, but according to our calculation, the sum over all (a,b,c) with triangle conditions is 17/21. However, if we compute the sum over all (a, b, c) without restrictions, it would be:

S_total = [(sum_{a=1}^∞ 2^a) * (sum_{b=1}^∞ 1/3^b) * (sum_{c=1}^∞ 1/5^c)].

But since sum_{a=1}^∞ 2^a diverges, but our S is finite. Therefore, our calculation of S = 17/21 must have been based on the constraints of the triangle conditions, which effectively limited the sums in such a way that convergence is achieved.

Alternatively, since we transformed the problem into a double sum over a and b, and then handled the inner sum over c, which was finite, and then the entire thing became convergent, it makes sense.

Therefore, unless there was a miscalculation in the steps, 17/21 might be the correct answer.

But let me check my calculation for S1 again:

For the first sum, we had Sum_{a=1}^∞ [ (2/3)^a * (5/2)(1 - (3/5)^a) ].

Which broke into 5/2 * Sum (2/3)^a - 5/2 * Sum (2/3)^a * (3/5)^a.

First term: 5/2 * [ (2/3) / (1 - 2/3) ] = 5/2 * [ (2/3) / (1/3) ] = 5/2 * 2 = 5. Correct.

Second term: 5/2 * Sum_{a=1}^∞ (2/3 * 3/5)^a = 5/2 * Sum_{a=1}^∞ (2/5)^a = 5/2 * [ (2/5) / (1 - 2/5) ] = 5/2 * [ (2/5) / (3/5) ] = 5/2 * 2/3 = 5/3. Correct.

Therefore, the first sum is 5 - 5/3 = 10/3. Then the second sum was 1/7. Therefore, total double sum is 10/3 + 1/7 = 73/21. Then S1 = 73/21 * 1/5 = 73/105. Then S2 = 1/21. Therefore, difference is 73/105 - 5/105 = 68/105. Multiply by 5/4: 68/105 * 5/4 = (68 * 5) / (105 * 4) = (340) / 420 = 17/21. Correct.

Therefore, I think this is correct.

But to be absolutely sure, let me check another way.

Suppose we think about generating functions again. The sum we computed, S = 17/21. Let me check if there is a way to get this result via another method.

Alternatively, note that the answer is 17/21. However, the problem says "Compute the sum...", and given that the logic checks out, and the steps are consistent, I think that 17/21 is the correct answer.

Therefore, the final answer is \boxed{\dfrac{17}{21}}.

**Final Answer**
\boxed{\dfrac{17}{21}}

To solve the problem, we are asked to compute the sum:

$$
\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c}
$$

where $ T $ is the set of all triples $ (a,b,c) $ of **positive integers** that can form the sides of a triangle. This requires that the triangle inequalities are satisfied:

$$
a + b > c, \quad a + c > b, \quad b + c > a.
$$

Alternatively, for a triangle with sides $ a, b, c $, the **largest side** must be **less than** the sum of the other two sides. So, if we denote $ M = \max(a,b,c) $, the triangle inequality is equivalent to:

$$
2M < a + b + c.
$$

This gives us a way to **redefine the sum** over all valid triples $ (a,b,c) $ by iterating over possible values of $ M $, and for each such $ M $, summing over the combinations $ (a,b,c) $ where $ \max(a,b,c) = M $ and $ a + b + c > 2M $.

---

### Step 1: Expressing the Sum

Let’s focus on computing the inner sum over $ c $. For each fixed $ a $ and $ b $, the values of $ c $ for which $ (a,b,c) $ form a triangle are those such that:

$$
\max(1, |a-b|+1) \le c \le a + b - 1.
$$

The number of such $ c $ values is:

$$
(a + b - 1) - \max(1, |a - b| + 1) + 1 = a + b - \max(1, |a - b| + 1).
$$

But instead of counting the number of $ c $, we are summing the term $ \frac{1}{5^c} $ over all valid $ c $ values, and then multiplying by $ \frac{2^a}{3^b} $.

So the full sum becomes:

$$
S = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b} \sum_{c=\max(1, |a-b| + 1)}^{a+b-1} \frac{1}{5^c}
$$

This is the core of the sum, and we aim to compute it.

---

### Step 2: Compute the Inner Sum

Let’s define the inner sum (from the earlier geometric sum) as:

$$
\sum_{c=L}^{U} \frac{1}{5^c} = \frac{1}{5^L} \cdot \frac{1 - \left(\frac{1}{5}\right)^{U-L+1}}{1 - \frac{1}{5}} = \frac{5}{4} \left( \frac{1}{5^L} - \frac{1}{5^{U+1}} \right)
$$

Now, for each $ a, b $, let:

- $ L = \max(1, |a-b| + 1) $
- $ U = a + b - 1 $

So the inner sum is:

$$
\sum_{c=L}^{U} \frac{1}{5^c} = \frac{5}{4} \left( \frac{1}{5^L} - \frac{1}{5^{U+1}} \right)
$$

Then the total sum becomes:

$$
S = \frac{5}{4} \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b} \left( \frac{1}{5^L} - \frac{1}{5^{U+1}} \right)
$$

This expression can be split into two parts:

$$
S = \frac{5}{4} \left( S_1 - S_2 \right)
$$

Where:

- $ S_1 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b \cdot 5^L} $
- $ S_2 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b \cdot 5^{U+1}} $

---

### Step 3: Computing $ S_2 $

Since $ U+1 = a+b $, the second inner sum becomes:

$$
S_2 = \sum_{a,b} \frac{2^a}{3^b} \cdot \frac{1}{5^{a+b}} = \left( \sum_{a=1}^\infty \left(\frac{2}{5} \right)^a \right) \cdot \left( \sum_{b=1}^\infty \left( \frac{1}{15} \right)^b \right)
$$

Compute these two geometric series:

$$
\sum_{a=1}^\infty \left( \frac{2}{5} \right)^a = \frac{2/5}{1 - 2/5} = \frac{2}{3}, \quad
\sum_{b=1}^\infty \left( \frac{1}{15} \right)^b = \frac{1/15}{1 - 1/15} = \frac{1}{14}
$$

So:

$$
S_2 = \frac{2}{3} \cdot \frac{1}{14} = \frac{1}{21}
$$

---

### Step 4: Computing $ S_1 $

$$
S_1 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b \cdot 5^{|a - b| + 1}} = \frac{1}{5} \sum_{a,b} \frac{2^a}{3^b} \cdot 5^{-|a - b|}
$$

Now, split the sum over $ a \geq b $ and $ a < b $, recognizing $ |a - b| $ is symmetric.

Let’s define:

- $ A = \sum_{a=1}^{\infty} \sum_{b=1}^{a} \frac{2^a}{3^b} \cdot 5^{-(a - b)} $
- $ B = \sum_{a=1}^{\infty} \sum_{b=a+1}^{\infty} \frac{2^a}{3^b} \cdot 5^{-(b - a)} $

Then $ S_1 = \frac{1}{5}(A + B) $

**Compute A:**

Let $ k = a - b \Rightarrow b = a - k, \; k = 0 $ to $ a-1 $

$$
A = \sum_{a=1}^\infty \sum_{k=0}^{a-1} \frac{2^a}{3^{a-k}} \cdot 5^{-k} = \sum_{a=1}^\infty \frac{2^a}{3^a} \sum_{k=0}^{a-1} \left( \frac{3}{5} \right)^k
$$

$$
\sum_{k=0}^{a-1} \left( \frac{3}{5} \right)^k = \frac{1 - (3/5)^a}{1 - 3/5} = \frac{5}{2} \left( 1 - \left( \frac{3}{5} \right)^a \right)
$$

So:

$$
A = \frac{5}{2} \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a \left(1 - \left( \frac{3}{5} \right)^a \right)
= \frac{5}{2} \left( \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a - \sum_{a=1}^\infty \left( \frac{2}{5} \right)^a \right)
= \frac{5}{2} \left(2 - \frac{2}{3} \cdot \frac{1}{1 - 2/5} \right) = \frac{10}{3}
$$

**Compute B:**

$$
B = \sum_{a=1}^\infty \sum_{m=1}^\infty \frac{2^a}{3^{a + m}} \cdot 5^{-m} = \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a \sum_{m=1}^\infty \left( \frac{1}{3 \cdot 5} \right)^m = \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a \cdot \frac{1}{14}
= \frac{1}{14} \cdot \frac{2}{3} \cdot \frac{1}{1 - 2/3} = \frac{1}{7}
$$

So:

$$
S_1 = \frac{1}{5} (A + B) = \frac{1}{5} \left( \frac{10}{3} + \frac{1}{7} \right) = \frac{1}{5} \cdot \frac{73}{21} = \frac{73}{105}
$$

---

### Step 5: Final Computation

We now have:

- $ S_1 = \frac{73}{105} $
- $ S_2 = \frac{1}{21} = \frac{5}{105} $

So:

$$
S = \frac{5}{4} (S_1 - S_2) = \frac{5}{4} \cdot \left( \frac{68}{105} \right) = \frac{340}{420} = \frac{17}{21}
$$

---

### Final Answer

$$
\boxed{\frac{17}{21}}
$$

[Assistant]
Explanation:
Okay, so I need to compute the sum over all triples (a, b, c) of positive integers that can form a triangle with sides a, b, c, of the term 2^a / (3^b * 5^c). Hmm. Let me think about how to approach this.

First, let's recall that for three positive integers a, b, c to form a triangle, they must satisfy the triangle inequalities. That is, the sum of any two sides must be greater than the third side. So, for a, b, c, we need:

a + b > c,
a + c > b,
b + c > a.

But since a, b, c are positive integers, maybe we can find a way to express these inequalities in terms of each variable being less than the sum of the other two. Alternatively, maybe we can fix an order to some extent. For example, without loss of generality, assume that a ≤ b ≤ c. Then, the triangle inequality reduces to a + b > c, since the other inequalities would automatically hold. For example, if a ≤ b ≤ c, then a + c > b is always true because c ≥ b and a is positive, so a + c ≥ a + b ≥ b + 1 > b. Similarly, b + c > a is also always true because b and c are both at least a, so their sum is at least 2a ≥ a. So the only non-trivial condition is a + b > c when a ≤ b ≤ c.

Therefore, maybe I can compute the sum by considering all triples (a, b, c) with a ≤ b ≤ c and a + b > c, then multiply by some symmetry factor? Wait, but the original problem doesn't specify any ordering, so each triple (a, b, c) can be in any order. However, since the variables a, b, c are in the exponents of different bases (2^a, 3^b, 5^c), the terms aren't symmetric. So permutations of (a, b, c) would lead to different terms. Therefore, maybe it's better to think of the problem as considering all triples where the triangle inequalities hold, regardless of order.

But that seems complicated. Maybe there's a generating function approach or inclusion-exclusion?

Alternatively, maybe I can think of the sum over all a, b, c ≥ 1 of 2^a / (3^b 5^c) where the triangle inequalities hold. So the total sum is equal to the sum over all triples (a, b, c) with a, b, c ≥ 1, of 2^a / (3^b 5^c) multiplied by an indicator function that is 1 if a, b, c can form a triangle, and 0 otherwise.

Therefore, the problem reduces to computing:

S = ∑_{a=1}^∞ ∑_{b=1}^∞ ∑_{c=1}^∞ [indicator(a, b, c satisfy triangle inequalities)] * (2^a / (3^b 5^c))

But how to compute this sum? It might be helpful to use the triangle inequalities and express the indicator as constraints on a, b, c. Since triangle inequalities are symmetric, maybe we can exploit some symmetry.

Alternatively, maybe we can compute the total sum over all a, b, c and subtract the ones that don't satisfy the triangle inequalities. That is:

Total sum without any restrictions is S_total = (∑_{a=1}^∞ 2^a) * (∑_{b=1}^∞ 1/3^b) * (∑_{c=1}^∞ 1/5^c).

But wait, but the original problem is not the total sum, but only the sum over triples that satisfy the triangle inequalities. So maybe:

S = S_total - S_bad, where S_bad is the sum over all triples that don't satisfy the triangle inequalities.

But how do I compute S_bad? Well, the triangle inequalities are not satisfied if at least one of the following is true:

a + b ≤ c,
a + c ≤ b,
b + c ≤ a.

So, S_bad is the sum over triples where at least one of these inequalities holds. But inclusion-exclusion might be needed here.

Alternatively, maybe we can compute S as follows:

First, compute S_total, then subtract the cases where the triangle inequalities are not satisfied. However, inclusion-exclusion is necessary because some triples might be counted multiple times if they fail more than one inequality.

So let me write that down.

Let me define:

S_total = (∑_{a=1}^∞ 2^a) * (∑_{b=1}^∞ 1/3^b) * (∑_{c=1}^∞ 1/5^c)

Compute each sum:

Sum_{a=1}^∞ 2^a = 2/(1 - 2) - 1? Wait, geometric series formula is sum_{n=0}^∞ r^n = 1/(1 - r). So sum from n=1 to ∞ is 1/(1 - r) - 1 = r/(1 - r). Therefore:

Sum_{a=1}^∞ 2^a = 2/(1 - 2) = -2? Wait, that's impossible, since sum_{a=1}^∞ 2^a diverges. Wait, but the original problem says that (a, b, c) are positive integers, so the term 2^a / (3^b 5^c) would have 2^a growing exponentially, but 3^b and 5^c are also exponential. However, the sum over all a, b, c might still diverge? Wait, but the problem says to compute the sum over all such triples. However, if the total sum diverges, then we have to be careful. However, since we need to compute only the sum over triples that satisfy the triangle inequalities, maybe the convergence is better?

Wait, let me check the convergence. Let me think about the total sum S_total:

Sum_{a=1}^∞ 2^a * Sum_{b=1}^∞ 1/3^b * Sum_{c=1}^∞ 1/5^c.

Each of these sums individually diverges? Wait, Sum_{a=1}^∞ 2^a is divergent. But in reality, the original problem must have a convergent sum because otherwise the problem would be ill-posed. However, since the term is 2^a / (3^b 5^c), which has different exponents for a, b, c. Wait, if we think of each term as 2^a * 3^{-b} * 5^{-c}, then the total sum is a product of three independent sums? Wait, no, the triangle inequalities impose some relation between a, b, c. However, without any constraints, the total sum is divergent because the 2^a term is divergent. Therefore, perhaps my approach is wrong?

But the problem states to compute the sum over all triples (a,b,c) that can form a triangle. However, if there are infinitely many such triples, and the terms don't go to zero fast enough, then the sum may diverge. But maybe due to the triangle condition, the sum converges? Hmm.

Wait, let me check for convergence. Suppose we fix a, b, c. The term is 2^a / (3^b 5^c). If we fix a and b, then as c increases, the term decreases as 1/5^c. However, for triangle inequalities, when we consider that for fixed a and b, c must be less than a + b. So for fixed a and b, the sum over c is sum_{c=1}^{a + b - 1} 1/5^c. Which converges as a finite sum. Similarly, if we fix other variables. Therefore, perhaps the total sum converges.

Wait, maybe I need to re-express the problem with the triangle inequalities.

But earlier, I thought about S_total, but maybe instead of considering all triples, I need to consider the sum with the triangle inequalities. Let me think again.

Let me think of the problem as first considering all a, b, c ≥ 1, and then subtracting those which don't satisfy triangle inequalities. However, since the total sum diverges (as sum_{a} 2^a is divergent), perhaps I need a different approach. Maybe instead of considering the entire sum, I can find a generating function that encodes the triangle conditions.

Alternatively, maybe I can fix an ordering and use symmetry. Earlier, I thought that if I assume a ≤ b ≤ c, then the triangle inequality reduces to a + b > c. Then, perhaps the original sum can be written as 6 times the sum over all a ≤ b ≤ c with a + b > c of the term 2^a / (3^b 5^c), plus some other terms? Wait, no, because the terms are not symmetric. Since the exponents are different for a, b, c, the permutations would give different terms. Therefore, we cannot just multiply by symmetry factors. For example, if a ≤ b ≤ c, then a, b, c could be permuted in different ways, but each permutation would have different values of 2^a / (3^b 5^c). Therefore, perhaps this approach is not straightforward.

Alternatively, let me try to consider the triangle inequalities in terms of variables. Let me think of the three triangle inequalities:

1. a + b > c
2. a + c > b
3. b + c > a

But since a, b, c are positive integers, we can think of the conditions as:

For all permutations, the sum of the two smaller sides must exceed the largest side.

Therefore, if I can find a way to count all triples (a, b, c) where the largest side is less than the sum of the other two. Therefore, perhaps if I consider for each triple, find which one is the largest. But again, the problem is the asymmetry in exponents.

Alternatively, maybe I can use generating functions with constraints.

Let me consider generating functions where I can encode the triangle inequalities.

But I need to think of variables a, b, c. Let me think about how to model the triangle inequalities.

Alternatively, perhaps I can fix the maximum value among a, b, c. Let's assume that c is the maximum. So c ≥ a, c ≥ b. Then, the triangle inequality requires that a + b > c. Similarly, if a is the maximum, then b + c > a, and if b is the maximum, then a + c > b. However, since the exponents are different, maybe it's better to consider each case where a particular variable is the maximum.

But again, since the exponents differ, this complicates things. Let me try.

Case 1: c is the largest side. So c ≥ a, c ≥ b, and a + b > c.

Case 2: b is the largest side. So b ≥ a, b ≥ c, and a + c > b.

Case 3: a is the largest side. So a ≥ b, a ≥ c, and b + c > a.

But since the triples (a, b, c) can be in any order, and the exponents differ, these cases will contribute different amounts to the sum. However, since the problem is symmetric with respect to permutations of a, b, c only if the exponents were the same. But here, since 2^a / (3^b 5^c) is not symmetric, we can't just multiply by the number of permutations. Therefore, perhaps we need to handle each case separately.

Alternatively, maybe we can think of the problem as follows:

For each triple (a, b, c) where the triangle inequalities hold, the term is 2^a / (3^b 5^c). So, if I can find a way to express this sum by considering for each possible maximum side, and then sum over all such possibilities.

Alternatively, maybe it's better to express the sum as the product of generating functions with constraints.

But I need to think of another way. Let me recall that the original problem is similar to counting the number of triangles with given side lengths, but with weights. So perhaps I need to find a way to express the conditions a + b > c, etc., in terms of generating functions.

Alternatively, think about the sum over all a, b, c ≥ 1 with a + b > c, a + c > b, and b + c > a. Let me try to express this as:

Sum_{a, b, c ≥ 1} [indicator(a + b > c and a + c > b and b + c > a)] * (2^a / (3^b 5^c)).

Which can be written as:

Sum_{a, b, c ≥ 1} (2^a / (3^b 5^c)) * [indicator(a + b > c) * indicator(a + c > b) * indicator(b + c > a)].

But how to compute this?

Alternatively, note that the triangle inequalities are equivalent to:

max(a, b, c) < a + b + c - max(a, b, c).

Wait, because for any triangle, the sum of the two smaller sides must exceed the largest side. Let me check. Suppose c is the largest side, then a + b > c. Since c > a and c > b, then a + b > c. But then the sum of all sides is a + b + c, so a + b = (a + b + c) - c. Therefore, the condition is that the largest side is less than the sum of the other two sides, which is equivalent to the largest side being less than half of the total perimeter.

Wait, more precisely, since if we denote S = a + b + c, then the largest side is less than S - largest side, so 2 * largest side < S. Therefore, largest side < S / 2. Therefore, the triangle inequalities are equivalent to the largest side being less than half of the total perimeter. Therefore, perhaps we can use this.

Therefore, if we let M = max(a, b, c), then the condition is M < (a + b + c)/2.

Alternatively, 2M < a + b + c.

But maybe this is helpful? Not sure yet.

Alternatively, since the problem is multiplicative, maybe I can express the sum as follows:

Let me think about fixing M, the maximum of a, b, c, and then sum over all triples where max(a, b, c) = M and the triangle inequalities hold. But again, the problem is the asymmetry in the exponents.

Alternatively, maybe I can think of generating functions where I consider the variables a, b, c, with constraints.

Wait, here's an idea. Let me consider generating functions for a, b, c with constraints. Let me define generating functions for each variable:

For variable a: it can be any positive integer, but with some conditions depending on b and c.

But since the triangle inequalities involve mutual relations between a, b, c, maybe it's complicated.

Alternatively, since we can consider the problem in terms of variables with ordering. Suppose we first fix an ordering, say a ≤ b ≤ c, then the triangle condition is a + b > c. Then, the total sum over all triples (a, b, c) can be expressed as the sum over all such ordered triples multiplied by the number of permutations that correspond to their different orderings. However, again, since the exponents are different, this complicates things.

Wait, perhaps not. Let's suppose that we consider ordered triples where a ≤ b ≤ c and a + b > c. Then, each such triple corresponds to certain permutations. However, because of the different exponents, the contribution of each permutation is different. For example, if we have a triple (a, b, c) with a < b < c, then the permutations where a is in different positions would have different exponents.

Therefore, maybe it's better to think that each triple (a, b, c) which satisfies the triangle inequalities can be in any order, but since the exponents are different, we need to consider each case.

Alternatively, maybe think of the original problem as:

Let me consider the sum over all a, b, c where a + b > c, a + c > b, and b + c > a. Since these are symmetric conditions, perhaps we can use some sort of generating function where we account for all three inequalities.

Alternatively, note that the three inequalities can be combined as:

For all i, j, k, with i, j, k being permutations of a, b, c: i + j > k.

But I don't know if that helps.

Alternatively, since each inequality is a + b > c, etc., we can note that for all triples (a, b, c), the triangle inequalities are equivalent to:

c < a + b,

b < a + c,

a < b + c.

But if we can express these as constraints on the variables. For example, if we fix c, then for a and b, we need a + b > c, and also a and b must be such that they don't violate other inequalities. Wait, but since a, b, c are symmetric except for the exponents, maybe we can fix c and consider the sum over a, b such that a + b > c, and then also considering the other inequalities? Wait, but if c is fixed, then the other inequalities would be a < b + c and b < a + c. Which are automatically satisfied if a and b are positive integers. Because a < b + c is equivalent to a - b < c. Since a and b are positive integers, unless a is very large compared to b and c. Wait, but for example, if c is fixed, and suppose a is very large, say a ≥ b + c + 1, then a < b + c would not hold. Therefore, that would be a case where a is too large. Therefore, so if we fix c, then to satisfy all triangle inequalities, we need:

a + b > c,

a < b + c,

b < a + c.

But if c is the largest side, then the first inequality is the only one needed. However, if c is not the largest side, then another inequality may be necessary.

This seems too vague. Maybe we need another approach.

Let me think about the following idea. Suppose I first consider all triples (a, b, c) with a, b, c ≥ 1, and then subtract those that do not satisfy the triangle conditions. However, as I thought earlier, the total sum diverges. However, perhaps the divergent parts cancel out?

Wait, but if the total sum diverges and the bad sum also diverges, then their difference might converge?

Alternatively, maybe there's another way. Let me think about the sum over all triples (a, b, c) with triangle inequalities. Since each triple must satisfy a + b > c, a + c > b, b + c > a. Let me try to find generating functions for each of these.

Alternatively, notice that the product of the three inequalities can be represented as follows:

For each triple (a, b, c), the triangle inequalities are equivalent to:

a + b + c > 2 * max(a, b, c).

Let me check this. Suppose that the maximum of a, b, c is M. Then the triangle inequalities are equivalent to M < a + b + c - M, which simplifies to 2M < a + b + c.

Therefore, the triangle inequalities are equivalent to 2 * max(a, b, c) < a + b + c. Therefore, perhaps we can use this condition to compute the sum.

Therefore, we can write the desired sum as:

Sum_{a, b, c ≥ 1} [indicator(2 * max(a, b, c) < a + b + c)] * (2^a / (3^b 5^c)).

Now, perhaps I can express this sum as follows: Let me consider cases depending on which variable is the maximum. Let me split the sum into three cases:

Case 1: a is the maximum, i.e., a ≥ b, a ≥ c.

Case 2: b is the maximum, i.e., b ≥ a, b ≥ c.

Case 3: c is the maximum, i.e., c ≥ a, c ≥ b.

Note that these cases are mutually exclusive and cover all possibilities where exactly one variable is the maximum. However, there are also cases where two variables are equal and the third is less, or all three are equal. But since the maximum can be tied (for example, a = b ≥ c), but in that case, we still have a single maximum (or two maxima). However, since we need to consider all triples where the maximum is less than the sum of the other two sides. Therefore, perhaps we can compute each case separately.

However, since the exponents are different for a, b, c, the contribution from each case is different. Therefore, I need to compute each case:

Let me define:

S1 = sum over a ≥ b, a ≥ c, and 2a < a + b + c of 2^a / (3^b 5^c)

S2 = sum over b ≥ a, b ≥ c, and 2b < a + b + c of 2^a / (3^b 5^c)

S3 = sum over c ≥ a, c ≥ b, and 2c < a + b + c of 2^a / (3^b 5^c)

Then, the total sum S = S1 + S2 + S3. However, we need to be careful not to double count cases where two variables are equal and the third is equal or less. But because of the max condition, each triple is counted exactly once in one of the cases. For example, if a = b = c, then all three cases would include this triple? Wait, no. For example, if all three are equal, then all three cases would satisfy the conditions. But actually, when considering case 1, a is the maximum, but if a = b = c, then a is equal to the other variables. However, in the definition of Case 1, it's "a ≥ b, a ≥ c". So if a = b = c, then the conditions are satisfied. But in this case, does it get counted in all three cases? No, because for each case, the maximum is considered. However, in reality, if all three variables are equal, then all three cases would include it? Wait, no. If we define S1 as a ≥ b, a ≥ c, then the triple (a, a, a) would be included in S1, S2, S3? Wait, no. For example, S1 is for when a is the maximum. If a = b = c, then a is the maximum. But also, b is the maximum and c is the maximum. So in reality, this triple would be counted in all three S1, S2, S3. Therefore, we would be overcounting. Therefore, this approach may not work unless we adjust for overlapping cases.

Therefore, to avoid overcounting, perhaps we should consider the cases where a, b, c are distinct, and then handle the equality cases separately. However, this complicates things further. Alternatively, perhaps it's better to use inclusion-exclusion, but I need to think.

Alternatively, maybe instead of splitting into cases where a, b, or c is the maximum, we can use the following approach. Let me consider that for each triple (a, b, c), the maximum of a, b, c is M. Then, the sum can be written as:

S = ∑_{M=1}^∞ ∑_{a,b,c with max(a,b,c)=M and 2M < a + b + c} (2^a / (3^b 5^c))

But even this seems complicated. However, perhaps for a given M, we can compute the sum over all triples (a, b, c) where max(a, b, c) = M and 2M < a + b + c.

But for each M, the condition max(a, b, c) = M implies that at least one of a, b, c is equal to M, and the others are ≤ M.

Furthermore, the triangle condition is 2M < a + b + c. Since max(a, b, c) = M, the sum a + b + c is at least M + 1 + 1 = M + 2 (if two other sides are 1) and at most 3M (if all sides are M). Therefore, the triangle condition 2M < a + b + c implies that a + b + c > 2M. Since a + b + c ≤ 3M, this condition becomes 2M < a + b + c ≤ 3M. Therefore, for each M, we have to sum over all triples (a, b, c) with max(a, b, c) = M and a + b + c > 2M.

But how can we compute this?

Let me fix M. Let me consider the sum over all (a, b, c) with max(a, b, c) = M and a + b + c > 2M.

First, let's compute the total number of triples with max(a, b, c) = M. Then, subtract those which have a + b + c ≤ 2M. However, since our problem is not about counting but about summing weighted terms, let me think.

But the problem is that the sum is over (a, b, c) with max(a, b, c) = M and a + b + c > 2M. So for each M, we need to compute:

Sum_{a, b, c, max(a,b,c)=M} [indicator(a + b + c > 2M)] * (2^a / (3^b 5^c))

But this seems challenging. However, perhaps I can find this sum by considering that for each M, the triples with max(a, b, c) = M can be partitioned into different cases depending on which variable is equal to M. Let me try:

Case 1: a = M, and b ≤ M, c ≤ M, and neither b nor c is equal to M. Then, the condition is max(a, b, c) = M.

Case 2: b = M, and a ≤ M, c ≤ M, and neither a nor c is equal to M.

Case 3: c = M, and a ≤ M, b ≤ M, and neither a nor b is equal to M.

Case 4: a = M, b = M, c < M.

Case 5: a = M, c = M, b < M.

Case 6: b = M, c = M, a < M.

Case 7: a = M, b = M, c = M.

But this is getting complicated, but maybe manageable.

However, since the exponents are different, perhaps we need to handle each case with its own contribution.

Alternatively, maybe we can think of all triples with max(a, b, c) = M. Let me note that the sum over all triples with max(a, b, c) = M is equal to:

Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M [indicator(max(a,b,c)=M)] * (2^a / (3^b 5^c))

But this is equal to:

Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M (2^a / (3^b 5^c)) * [indicator(max(a,b,c)=M)]

This can be rewritten as:

Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M (2^a / (3^b 5^c)) * [indicator(a = M or b = M or c = M) - indicator(a = M and b = M) - ... + ... ]

But inclusion-exclusion here is going to get messy. Alternatively, note that the number of triples where max(a,b,c)=M is equal to 3*(M^2) - 3*M + 1, but again with weights.

Alternatively, maybe we can use the principle that the sum over max(a,b,c)=M is equal to the sum over all triples with a, b, c ≤ M, and subtract those with max(a,b,c) < M. Therefore:

Sum_{max(a,b,c)=M} f(a,b,c) = Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M f(a,b,c) - Sum_{max(a,b,c) < M} f(a,b,c)

But the second term is Sum_{a=1}^{M-1} Sum_{b=1}^{M-1} Sum_{c=1}^{M-1} f(a,b,c)

Therefore, for our problem, the sum over all triples with max(a,b,c)=M and a + b + c > 2M is equal to:

[Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M f(a,b,c) * indicator(max(a,b,c)=M)] - [Sum_{a=1}^M Sum_{b=1}^M Sum_{c=1}^M f(a,b,c) * indicator(max(a,b,c)=M) * indicator(a + b + c ≤ 2M)]

But maybe it's better to compute it as:

For each M, the sum over triples with max(a,b,c) = M is:

Total = [sum_{a=1}^M 2^a] * [sum_{b=1}^M 1/3^b] * [sum_{c=1}^M 1/5^c] - [sum_{a=1}^{M-1} 2^a] * [sum_{b=1}^{M-1} 1/3^b] * [sum_{c=1}^{M-1} 1/5^c]

But this seems complex, but maybe manageable. However, we need to then subtract from this the sum over those triples where a + b + c ≤ 2M. Therefore, it's getting very involved.

Alternatively, maybe it's better to think of the original problem as follows. Since the triangle conditions are symmetric, perhaps I can use generating functions with constraints. Let me try.

Let me think of generating functions where I have variables a, b, c with a + b > c, a + c > b, b + c > a. Let me note that these inequalities can be rewritten as:

c < a + b,

b < a + c,

a < b + c.

But in generating functions, how do I encode these inequalities? Maybe consider generating functions for each variable, but with constraints.

Alternatively, for each variable, we can think of generating functions and use the constraints. Let me consider the generating function for a, b, c.

But I need to think of generating functions that satisfy a + b > c, etc. This seems challenging. However, note that these inequalities can be combined into the single condition that the sum of the two smaller sides is greater than the largest side, as I thought earlier.

Therefore, let me consider that the generating function for each triple is 2^a / (3^b 5^c) multiplied by the indicator that the sum of the two smaller sides is greater than the largest side.

But how to model this?

Alternatively, here's another idea. Since the triangle inequalities are equivalent to the condition that no one side is greater than or equal to the sum of the other two. Therefore, the complement of this is that at least one side is greater than or equal to the sum of the other two. Therefore, we can write the sum S as:

S = Total - S_bad,

where Total is the sum over all triples (a, b, c) of 2^a / (3^b 5^c), and S_bad is the sum over triples where at least one side is ≥ the sum of the other two. But as before, Total is divergent. So this might not help, but maybe the divergences cancel out?

Alternatively, maybe I can compute S_total and S_bad with some regularization? But that seems too vague.

Alternatively, think of the problem as follows: For variables a, b, c, the triangle inequalities can be written as:

c < a + b,

b < a + c,

a < b + c.

Which is equivalent to:

For all permutations, the side is less than the sum of the other two.

Alternatively, notice that if we consider the three inequalities:

c < a + b,

a < b + c,

b < a + c.

These are symmetric. Therefore, maybe I can consider the generating function where I have the product of generating functions for each variable, multiplied by constraints.

Alternatively, consider that the three inequalities can be rewritten as:

c < a + b => c ≤ a + b - 1,

Similarly for others. Therefore, if I can write the generating function as:

Sum_{a,b,c ≥ 1} 2^a / (3^b 5^c) * [indicator(c ≤ a + b - 1) * indicator(a ≤ b + c - 1) * indicator(b ≤ a + c - 1)]

But how to compute this?

Alternatively, note that the three inequalities can be combined into a single condition. Let me think about fixing a and b, then c must satisfy both c < a + b and c > |a - b|. Wait, because for the triangle inequality, if we fix a and b, then c must satisfy |a - b| < c < a + b. But since c is a positive integer, c must satisfy max(1, |a - b| + 1) ≤ c ≤ a + b - 1.

Wait, but this is only if we fix a and b. However, the other inequalities also need to be considered. Wait, if we fix a and b, then c needs to satisfy c < a + b, and also, since a and b are fixed, the other inequalities:

a < b + c => c > a - b.

But since c is positive, this implies c ≥ max(1, a - b + 1).

Similarly, b < a + c => c > b - a => c ≥ max(1, b - a + 1).

Therefore, combining all, for fixed a and b, c must satisfy:

max(1, |a - b| + 1) ≤ c ≤ a + b - 1.

But since c ≥ 1, and |a - b| + 1 is at least 1 (since |a - b| is non-negative). Therefore, for fixed a and b, the range of c is from max(1, |a - b| + 1) to a + b - 1.

However, note that if a and b are such that a + b - 1 ≥ 1, which is true for a, b ≥ 1, except when a = b = 1, in which case c must be between max(1, 0 + 1) = 1 and 1 + 1 - 1 = 1. So c = 1. Which is correct, because for a = b = 1, c must be < 2, so c = 1. Which is valid.

Therefore, this seems promising. Therefore, the original sum can be rewritten as:

S = ∑_{a=1}^∞ ∑_{b=1}^∞ [∑_{c = max(1, |a - b| + 1)}^{a + b - 1} 2^a / (3^b 5^c)]

Similarly, since for each a and b, c is determined to be in that range. Therefore, the problem reduces to computing this double sum over a and b, and for each, summing over c. So maybe we can compute this.

So, let me focus on this. Let me denote:

For each a and b, compute the inner sum over c. Let me denote:

For fixed a and b, the inner sum is:

Sum_{c = L}^{U} 1/5^c, where L = max(1, |a - b| + 1), and U = a + b - 1.

Therefore, the total sum S is equal to:

Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * [Sum_{c = L}^{U} 1/5^c]

Therefore, if I can compute this inner sum over c, then multiply by 2^a / 3^b, and then sum over a and b, I can get the desired result.

Okay, let's try to compute this inner sum. Let me first handle the inner sum.

First, note that Sum_{c = L}^{U} 1/5^c is a geometric series from c = L to c = U. The formula for this is (1/5^L - 1/5^{U + 1}) / (1 - 1/5)) = (1/5^L - 1/5^{U + 1}) / (4/5) ) = 5/4 * (1/5^L - 1/5^{U + 1}).

Therefore, the inner sum is 5/4 * (1/5^L - 1/5^{U + 1}).

Therefore, substituting back, S becomes:

S = (5/4) * Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * [1/5^L - 1/5^{U + 1}]

Where L = max(1, |a - b| + 1) and U = a + b - 1.

So, S = (5/4) * [Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * (1/5^L - 1/5^{U + 1}) ]

Therefore, let me split this into two sums:

S = (5/4)[ S1 - S2 ], where

S1 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^L,

and

S2 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{U + 1}.

So, let me compute S1 and S2 separately.

First, let's compute S1 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{L}, where L = max(1, |a - b| + 1).

Similarly, S2 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{U + 1}, where U = a + b - 1, so U + 1 = a + b.

Therefore, S2 becomes Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{a + b} = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 5^a) * (1/3^b / 5^b) ) = Sum_{a=1}^∞ (2/5)^a * Sum_{b=1}^∞ (1/15)^b.

Compute this:

Sum_{a=1}^∞ (2/5)^a = (2/5) / (1 - 2/5) = (2/5) / (3/5) = 2/3.

Sum_{b=1}^∞ (1/15)^b = (1/15) / (1 - 1/15) = (1/15) / (14/15) = 1/14.

Therefore, S2 = (2/3) * (1/14) = 2/(42) = 1/21.

Okay, so S2 is 1/21. Great, that was straightforward.

Now, S1 is more complicated: Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1/5^{max(1, |a - b| + 1)}.

Let me think about how to handle this. The term max(1, |a - b| + 1) can be simplified. Let me note that |a - b| + 1 is always at least 1, since |a - b| ≥ 0. Therefore, max(1, |a - b| + 1) is just |a - b| + 1. Wait, no! Wait: if |a - b| + 1 ≥ 1, which it always is, because |a - b| ≥ 0, so |a - b| + 1 ≥ 1. Therefore, actually, L = max(1, |a - b| + 1) = |a - b| + 1. Wait, that seems contradictory to earlier reasoning. Wait, no, let me check. The earlier idea was that for fixed a and b, c must be ≥ max(1, |a - b| + 1). However, |a - b| + 1 is always ≥ 1 since |a - b| ≥ 0. Therefore, max(1, |a - b| + 1) is just |a - b| + 1. Therefore, the inner sum over c becomes from c = |a - b| + 1 to c = a + b - 1. Wait, but if |a - b| + 1 is equal to L, then why was there a max(1, ...)? Because if |a - b| + 1 could be less than 1, but since |a - b| ≥ 0, then |a - b| + 1 ≥ 1. Therefore, L = |a - b| + 1.

Therefore, S1 is Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1 / 5^{ |a - b| + 1 }.

So S1 = (1/5) * Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) / 5^{|a - b|}.

Simplify: S1 = (1/5) * Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 5^{-|a - b|}.

So, this seems like a double sum over a and b of (2^a / 3^b) * 5^{-|a - b|}. Let me try to compute this.

To compute this double sum, perhaps I can split it into cases where a ≥ b and a < b. Since |a - b| is symmetric.

Let me denote for a ≥ b, |a - b| = a - b.

For a < b, |a - b| = b - a.

Therefore, the double sum becomes:

Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 5^{-|a - b|} = Sum_{a=1}^∞ Sum_{b=1}^a (2^a / 3^b) * 5^{-(a - b)} + Sum_{a=1}^∞ Sum_{b=a+1}^∞ (2^a / 3^b) * 5^{-(b - a)}.

Let me handle these two sums separately.

First, let's compute the first sum: Sum_{a=1}^∞ Sum_{b=1}^a (2^a / 3^b) * 5^{-(a - b)}.

Let me change variables. Let k = a - b. Since b ranges from 1 to a, k ranges from 0 to a - 1. So for each a, we can write:

Sum_{b=1}^a (2^a / 3^b) * 5^{-k} where k = a - b.

But k = a - b => b = a - k.

Therefore, substituting:

Sum_{k=0}^{a - 1} (2^a / 3^{a - k}) * 5^{-k} = 2^a * Sum_{k=0}^{a - 1} (1/3^{a - k}) * (1/5^k) = 2^a * (1/3^a) * Sum_{k=0}^{a - 1} (3/5)^k.

Wait, let me check:

Since 1 / 3^{a - k} = 3^k / 3^a, so:

(1/3^{a - k}) * (1/5^k) = 3^k / 3^a * 1 / 5^k = (3/5)^k / 3^a. Therefore:

Sum_{k=0}^{a - 1} (3/5)^k / 3^a = (1 / 3^a) * Sum_{k=0}^{a - 1} (3/5)^k.

Therefore, the entire first term becomes:

Sum_{a=1}^∞ [ 2^a / 3^a * Sum_{k=0}^{a - 1} (3/5)^k ].

Similarly, let's compute the second sum: Sum_{a=1}^∞ Sum_{b=a+1}^∞ (2^a / 3^b) * 5^{-(b - a)}.

Here, let me change variables again. Let m = b - a. Since b ≥ a + 1, m ≥ 1. So for each a, b = a + m, m ≥ 1.

Then:

Sum_{a=1}^∞ Sum_{m=1}^∞ (2^a / 3^{a + m}) * 5^{-m} = Sum_{a=1}^∞ 2^a / 3^a * Sum_{m=1}^∞ (1 / 3^m) * (1 / 5^m) ) = Sum_{a=1}^∞ (2/3)^a * Sum_{m=1}^∞ (1 / 15)^m.

Because 1 / 3^m * 1 / 5^m = 1 / 15^m.

Therefore, the second sum is:

Sum_{a=1}^∞ (2/3)^a * [ (1/15) / (1 - 1/15) ) ] = Sum_{a=1}^∞ (2/3)^a * [ 1 / 14 ] = (1 / 14) * [ (2/3) / (1 - 2/3) ) ] = (1 / 14) * [ (2/3) / (1/3) ) ] = (1 / 14) * 2 = 2 / 14 = 1 / 7.

Okay, so the second sum is 1/7. Now, going back to the first sum:

First sum is Sum_{a=1}^∞ [ (2/3)^a * Sum_{k=0}^{a - 1} (3/5)^k ].

Let me compute this. Let me note that Sum_{k=0}^{a - 1} (3/5)^k is a finite geometric series. The sum is [1 - (3/5)^a] / [1 - 3/5] = [1 - (3/5)^a] / (2/5) ) = (5/2)[1 - (3/5)^a].

Therefore, the first sum becomes:

Sum_{a=1}^∞ (2/3)^a * (5/2)[1 - (3/5)^a] = (5/2) Sum_{a=1}^∞ (2/3)^a - (5/2) Sum_{a=1}^∞ (2/3)^a * (3/5)^a.

Compute each part:

First term: (5/2) * Sum_{a=1}^∞ (2/3)^a = (5/2) * [ (2/3) / (1 - 2/3) ) ] = (5/2) * [ (2/3) / (1/3) ) ] = (5/2) * 2 = 5.

Second term: (5/2) * Sum_{a=1}^∞ (2/3 * 3/5)^a = (5/2) * Sum_{a=1}^∞ (2/5)^a = (5/2) * [ (2/5) / (1 - 2/5) ) ] = (5/2) * [ (2/5) / (3/5) ) ] = (5/2) * (2/3) = 5/3.

Therefore, the first sum is 5 - 5/3 = 10/3.

Therefore, combining both sums, the total double sum (for S1) is 10/3 + 1/7 = (70/21 + 3/21) = 73/21.

Wait, no: the first sum was 10/3 and the second sum was 1/7, so total double sum is 10/3 + 1/7 = (70 + 3)/21 = 73/21.

Therefore, S1 = (1/5) * 73/21 = 73 / 105.

Therefore, S = (5/4)(S1 - S2) = (5/4)(73/105 - 1/21).

Compute 73/105 - 1/21:

Convert 1/21 to 5/105, so 73/105 - 5/105 = 68/105.

Then, S = (5/4)(68/105) = (5 * 68) / (4 * 105) = (340) / 420 = Simplify numerator and denominator by 20: 17 / 21. Wait:

Wait, 5*68 = 340, 4*105 = 420.

Divide numerator and denominator by 20: 340 / 420 = 17 / 21. Yes.

Therefore, S = 17/21.

Wait, that seems too clean. Let me check the calculations again.

First, S1 was calculated as (1/5) * 73/21, which is 73/105. Then S2 is 1/21. So S1 - S2 is 73/105 - 5/105 = 68/105. Then multiply by 5/4: (68/105) * (5/4) = (68 * 5) / (105 * 4) = 340 / 420 = simplify by dividing numerator and denominator by 20: 17 / 21. Yes, that's correct.

So the final answer is 17/21?

Wait, but let me check again the steps for errors.

First, S2 was calculated as 1/21. Let me confirm:

For S2, we had:

S2 = Sum_{a=1}^∞ Sum_{b=1}^∞ (2^a / 3^b) * 1 / 5^{a + b}.

This is equal to [Sum_{a=1}^∞ (2/5)^a] * [Sum_{b=1}^∞ (1/15)^b].

Wait, hold on. Wait, 1/5^{a + b} = 1/5^a * 1/5^b. Therefore, the double sum becomes [Sum_{a} (2^a / 5^a)] * [Sum_b (1 / 3^b / 5^b)] ?

Wait, no. Wait, original expression:

(2^a / 3^b) * 1 / 5^{a + b} = (2^a / 5^a) * (1 / 3^b / 5^b) ?

Wait, no. It's (2^a / 3^b) * 1 / 5^{a + b} = 2^a / (3^b * 5^{a + b}) ) = (2 / 5)^a * (1 / (3 * 5)^b) ) = (2/5)^a * (1/15)^b. Therefore, the sum is [Sum_{a=1}^∞ (2/5)^a] * [Sum_{b=1}^∞ (1/15)^b]. Which is (2/3) * (1/14) = 1/21. Correct.

Then for S1, we had that the double sum became 73/21 and then multiplied by 1/5 gives 73/105. Then S1 - S2 = 73/105 - 5/105 = 68/105. Multiply by 5/4 gives 17/21. Seems correct.

But the problem says to compute the sum over all triples (a,b,c) that can form a triangle of 2^a / (3^b 5^c). So according to this calculation, the answer is 17/21. Is that possible?

But let me check if the logic is correct. When I broke the sum into S1 and S2, and then calculated them, the steps seem valid. But to be thorough, let me check with small values.

Alternatively, let me think about the total sum S_total. Earlier, we thought that it diverges, but in our calculation, we subtracted S2 from S1 and got a finite value. However, since both S1 and S2 were calculated as finite sums (due to the constraints on c), maybe it works out. Also, since we transformed the original problem into a convergent sum, it's possible.

Therefore, maybe the answer is 17/21. However, let me check with a simple case.

For example, consider all possible small triples (a, b, c) and compute the sum manually, then compare with the result.

But since the answer is a fraction, and we arrived at 17/21, let me see if there is any chance of error in the steps.

First, when calculating S1, we split into a ≥ b and a < b. Then converted the first sum into a double sum over a and k, etc. Then we found the first sum as 10/3 and the second sum as 1/7, totaling 73/21. Then S1 is that times 1/5: 73/105. Then S2 is 1/21. Then (S1 - S2) * 5/4 = 17/21. Seems correct.

Alternatively, perhaps I can check with M = 1, compute the contribution.

When M = 1, the only possible (a, b, c) with max(a, b, c) = 1 is (1,1,1). The triangle inequality condition is 1 + 1 > 1, which holds. Therefore, the contribution is 2^1 / (3^1 5^1) = 2/(3*5) = 2/15.

But according to our formula, let's check what part of the overall sum comes from M=1. Wait, but our derivation was based on a different approach, so maybe it's not straightforward.

Alternatively, since our final answer is 17/21 ≈ 0.8095, which is less than 1, which seems plausible.

Alternatively, let me think of the total sum over all (a,b,c) without triangle conditions. Wait, earlier we thought it was divergent, but according to our calculation, the sum over all (a,b,c) with triangle conditions is 17/21. However, if we compute the sum over all (a, b, c) without restrictions, it would be:

S_total = [(sum_{a=1}^∞ 2^a) * (sum_{b=1}^∞ 1/3^b) * (sum_{c=1}^∞ 1/5^c)].

But since sum_{a=1}^∞ 2^a diverges, but our S is finite. Therefore, our calculation of S = 17/21 must have been based on the constraints of the triangle conditions, which effectively limited the sums in such a way that convergence is achieved.

Alternatively, since we transformed the problem into a double sum over a and b, and then handled the inner sum over c, which was finite, and then the entire thing became convergent, it makes sense.

Therefore, unless there was a miscalculation in the steps, 17/21 might be the correct answer.

But let me check my calculation for S1 again:

For the first sum, we had Sum_{a=1}^∞ [ (2/3)^a * (5/2)(1 - (3/5)^a) ].

Which broke into 5/2 * Sum (2/3)^a - 5/2 * Sum (2/3)^a * (3/5)^a.

First term: 5/2 * [ (2/3) / (1 - 2/3) ] = 5/2 * [ (2/3) / (1/3) ] = 5/2 * 2 = 5. Correct.

Second term: 5/2 * Sum_{a=1}^∞ (2/3 * 3/5)^a = 5/2 * Sum_{a=1}^∞ (2/5)^a = 5/2 * [ (2/5) / (1 - 2/5) ] = 5/2 * [ (2/5) / (3/5) ] = 5/2 * 2/3 = 5/3. Correct.

Therefore, the first sum is 5 - 5/3 = 10/3. Then the second sum was 1/7. Therefore, total double sum is 10/3 + 1/7 = 73/21. Then S1 = 73/21 * 1/5 = 73/105. Then S2 = 1/21. Therefore, difference is 73/105 - 5/105 = 68/105. Multiply by 5/4: 68/105 * 5/4 = (68 * 5) / (105 * 4) = (340) / 420 = 17/21. Correct.

Therefore, I think this is correct.

But to be absolutely sure, let me check another way.

Suppose we think about generating functions again. The sum we computed, S = 17/21. Let me check if there is a way to get this result via another method.

Alternatively, note that the answer is 17/21. However, the problem says "Compute the sum...", and given that the logic checks out, and the steps are consistent, I think that 17/21 is the correct answer.

Therefore, the final answer is \boxed{\dfrac{17}{21}}.

**Final Answer**
\boxed{\dfrac{17}{21}}

To solve the problem, we are asked to compute the sum:

$$
\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c}
$$

where $ T $ is the set of all triples $ (a,b,c) $ of **positive integers** that can form the sides of a triangle. This requires that the triangle inequalities are satisfied:

$$
a + b > c, \quad a + c > b, \quad b + c > a.
$$

Alternatively, for a triangle with sides $ a, b, c $, the **largest side** must be **less than** the sum of the other two sides. So, if we denote $ M = \max(a,b,c) $, the triangle inequality is equivalent to:

$$
2M < a + b + c.
$$

This gives us a way to **redefine the sum** over all valid triples $ (a,b,c) $ by iterating over possible values of $ M $, and for each such $ M $, summing over the combinations $ (a,b,c) $ where $ \max(a,b,c) = M $ and $ a + b + c > 2M $.

---

### Step 1: Expressing the Sum

Let’s focus on computing the inner sum over $ c $. For each fixed $ a $ and $ b $, the values of $ c $ for which $ (a,b,c) $ form a triangle are those such that:

$$
\max(1, |a-b|+1) \le c \le a + b - 1.
$$

The number of such $ c $ values is:

$$
(a + b - 1) - \max(1, |a - b| + 1) + 1 = a + b - \max(1, |a - b| + 1).
$$

But instead of counting the number of $ c $, we are summing the term $ \frac{1}{5^c} $ over all valid $ c $ values, and then multiplying by $ \frac{2^a}{3^b} $.

So the full sum becomes:

$$
S = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b} \sum_{c=\max(1, |a-b| + 1)}^{a+b-1} \frac{1}{5^c}
$$

This is the core of the sum, and we aim to compute it.

---

### Step 2: Compute the Inner Sum

Let’s define the inner sum (from the earlier geometric sum) as:

$$
\sum_{c=L}^{U} \frac{1}{5^c} = \frac{1}{5^L} \cdot \frac{1 - \left(\frac{1}{5}\right)^{U-L+1}}{1 - \frac{1}{5}} = \frac{5}{4} \left( \frac{1}{5^L} - \frac{1}{5^{U+1}} \right)
$$

Now, for each $ a, b $, let:

- $ L = \max(1, |a-b| + 1) $
- $ U = a + b - 1 $

So the inner sum is:

$$
\sum_{c=L}^{U} \frac{1}{5^c} = \frac{5}{4} \left( \frac{1}{5^L} - \frac{1}{5^{U+1}} \right)
$$

Then the total sum becomes:

$$
S = \frac{5}{4} \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b} \left( \frac{1}{5^L} - \frac{1}{5^{U+1}} \right)
$$

This expression can be split into two parts:

$$
S = \frac{5}{4} \left( S_1 - S_2 \right)
$$

Where:

- $ S_1 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b \cdot 5^L} $
- $ S_2 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b \cdot 5^{U+1}} $

---

### Step 3: Computing $ S_2 $

Since $ U+1 = a+b $, the second inner sum becomes:

$$
S_2 = \sum_{a,b} \frac{2^a}{3^b} \cdot \frac{1}{5^{a+b}} = \left( \sum_{a=1}^\infty \left(\frac{2}{5} \right)^a \right) \cdot \left( \sum_{b=1}^\infty \left( \frac{1}{15} \right)^b \right)
$$

Compute these two geometric series:

$$
\sum_{a=1}^\infty \left( \frac{2}{5} \right)^a = \frac{2/5}{1 - 2/5} = \frac{2}{3}, \quad
\sum_{b=1}^\infty \left( \frac{1}{15} \right)^b = \frac{1/15}{1 - 1/15} = \frac{1}{14}
$$

So:

$$
S_2 = \frac{2}{3} \cdot \frac{1}{14} = \frac{1}{21}
$$

---

### Step 4: Computing $ S_1 $

$$
S_1 = \sum_{a=1}^{\infty} \sum_{b=1}^{\infty} \frac{2^a}{3^b \cdot 5^{|a - b| + 1}} = \frac{1}{5} \sum_{a,b} \frac{2^a}{3^b} \cdot 5^{-|a - b|}
$$

Now, split the sum over $ a \geq b $ and $ a < b $, recognizing $ |a - b| $ is symmetric.

Let’s define:

- $ A = \sum_{a=1}^{\infty} \sum_{b=1}^{a} \frac{2^a}{3^b} \cdot 5^{-(a - b)} $
- $ B = \sum_{a=1}^{\infty} \sum_{b=a+1}^{\infty} \frac{2^a}{3^b} \cdot 5^{-(b - a)} $

Then $ S_1 = \frac{1}{5}(A + B) $

**Compute A:**

Let $ k = a - b \Rightarrow b = a - k, \; k = 0 $ to $ a-1 $

$$
A = \sum_{a=1}^\infty \sum_{k=0}^{a-1} \frac{2^a}{3^{a-k}} \cdot 5^{-k} = \sum_{a=1}^\infty \frac{2^a}{3^a} \sum_{k=0}^{a-1} \left( \frac{3}{5} \right)^k
$$

$$
\sum_{k=0}^{a-1} \left( \frac{3}{5} \right)^k = \frac{1 - (3/5)^a}{1 - 3/5} = \frac{5}{2} \left( 1 - \left( \frac{3}{5} \right)^a \right)
$$

So:

$$
A = \frac{5}{2} \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a \left(1 - \left( \frac{3}{5} \right)^a \right)
= \frac{5}{2} \left( \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a - \sum_{a=1}^\infty \left( \frac{2}{5} \right)^a \right)
= \frac{5}{2} \left(2 - \frac{2}{3} \cdot \frac{1}{1 - 2/5} \right) = \frac{10}{3}
$$

**Compute B:**

$$
B = \sum_{a=1}^\infty \sum_{m=1}^\infty \frac{2^a}{3^{a + m}} \cdot 5^{-m} = \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a \sum_{m=1}^\infty \left( \frac{1}{3 \cdot 5} \right)^m = \sum_{a=1}^\infty \left( \frac{2}{3} \right)^a \cdot \frac{1}{14}
= \frac{1}{14} \cdot \frac{2}{3} \cdot \frac{1}{1 - 2/3} = \frac{1}{7}
$$

So:

$$
S_1 = \frac{1}{5} (A + B) = \frac{1}{5} \left( \frac{10}{3} + \frac{1}{7} \right) = \frac{1}{5} \cdot \frac{73}{21} = \frac{73}{105}
$$

---

### Step 5: Final Computation

We now have:

- $ S_1 = \frac{73}{105} $
- $ S_2 = \frac{1}{21} = \frac{5}{105} $

So:

$$
S = \frac{5}{4} (S_1 - S_2) = \frac{5}{4} \cdot \left( \frac{68}{105} \right) = \frac{340}{420} = \frac{17}{21}
$$

---

### Final Answer

$$
\boxed{\frac{17}{21}}
$$

Solution:
For a triangle with side lengths $a,$ $b,$ $c,$ let $s = \frac{a + b + c}{2},$ and let
\begin{align*}
x &= s - a = \frac{-a + b + c}{2}, \\
y &= s - b = \frac{a - b + c}{2}, \\
z &= s - c = \frac{a + b - c}{2}.
\end{align*}By the Triangle Inequality, $x,$ $y,$ and $z$ are all positive.  (This technique is often referred to as the Ravi Substitution.)  Note that
\begin{align*}
a &= y + z, \\
b &= x + z, \\
c &= x + y.
\end{align*}If $s$ is even, then $x,$ $y,$ and $z$ are all positive integers.  So, we can set $x = i,$ $y = j,$ and $z = k,$ which gives us the parameterization $(a,b,c) = (j + k, i + k, i + j).$

If $s$ is odd, then $x,$ $y,$ and $z$ are all of the form $n - \frac{1}{2},$ where $n$ is a positive integer.  So, we can set $x = i - \frac{1}{2},$ $y = j - \frac{1}{2},$ and $z = k - \frac{1}{2}.$  This gives us the parameterization $(a,b,c) = (j + k - 1, i + k - 1, i + j - 1).$

Thus, our sum is
\begin{align*}
\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c} &= \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \left( \frac{2^{j + k}}{3^{i + k} 5^{i + j}} + \frac{2^{j + k - 1}}{3^{i + k - 1} 5^{i + j - 1}} \right) \\
&= \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \left( \frac{2^{j + k}}{3^{i + k} 5^{i + j}} + \frac{15}{2} \cdot \frac{2^{j + k}}{3^{i + k} 5^{i + j}} \right) \\
&= \frac{17}{2} \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{2^{j + k}}{3^{i + k} 5^{i + j}} \\
&= \frac{17}{2} \sum_{i = 1}^\infty \frac{1}{15^i} \sum_{j = 1}^\infty \left( \frac{2}{5} \right)^j \sum_{k = 1}^\infty \left( \frac{2}{3} \right)^k \\
&= \frac{17}{2} \cdot \frac{1/15}{1 - 1/15} \cdot \frac{2/5}{1 - 2/5} \cdot \frac{2/3}{1 - 2/3} \\
&= \boxed{\frac{17}{21}}.
\end{align*}