Okay, so I need to find the value of f(2015, 2) based on the given recursive function. Let me try to understand the function first. The function f is defined with three cases:

1. If i = 0 and 0 ≤ j ≤ 4, then f(i, j) is mod5(j + 1). That means for the base case when i is 0, we just take j + 1 modulo 5. So, for example, f(0, 0) = 1, f(0, 1) = 2, up to f(0, 4) = 5 mod5 = 0.

2. If i ≥ 1 and j = 0, then f(i, j) is f(i - 1, 1). So when we're at any level i (greater than 0) and j is 0, we look one step up in i and set j to 1. So f(1, 0) = f(0, 1) = 2, f(2, 0) = f(1, 1), and so on.

3. If i ≥ 1 and 1 ≤ j ≤ 4, then f(i, j) is f(i - 1, f(i, j - 1)). This is a recursive step where we have to compute f(i, j - 1) first, then use that result as the new j in the previous level of i. So this is like a nested recursion, which might get complicated, but perhaps there's a pattern here.

The question is asking for f(2015, 2). 2015 is a pretty large number, so directly computing it step by step might not be feasible. I need to find a pattern or some periodicity in the function's behavior, maybe modulo some number, which could help simplify the calculation.

Let me start by computing some small values of f(i, j) to see if I can spot a pattern.

First, let's build a table for small i and j. Let's start with i = 0 to, say, 4, and j = 0 to 4.

For i = 0:
- f(0, 0) = mod5(0 + 1) = 1
- f(0, 1) = mod5(1 + 1) = 2
- f(0, 2) = mod5(2 + 1) = 3
- f(0, 3) = mod5(3 + 1) = 4
- f(0, 4) = mod5(4 + 1) = 0

So the first row is 1, 2, 3, 4, 0.

Now, let's compute i = 1.

For i = 1, j = 0: f(1, 0) = f(0, 1) = 2
For i = 1, j = 1: f(1, 1) = f(0, f(1, 0)) = f(0, 2) = 3
For i = 1, j = 2: f(1, 2) = f(0, f(1, 1)) = f(0, 3) = 4
For i = 1, j = 3: f(1, 3) = f(0, f(1, 2)) = f(0, 4) = 0
For i = 1, j = 4: f(1, 4) = f(0, f(1, 3)) = f(0, 0) = 1

So the second row is 2, 3, 4, 0, 1.

Now, i = 2:

j = 0: f(2, 0) = f(1, 1) = 3
j = 1: f(2, 1) = f(1, f(2, 0)) = f(1, 3) = 0
j = 2: f(2, 2) = f(1, f(2, 1)) = f(1, 0) = 2
j = 3: f(2, 3) = f(1, f(2, 2)) = f(1, 2) = 4
j = 4: f(2, 4) = f(1, f(2, 3)) = f(1, 4) = 1

So the third row is 3, 0, 2, 4, 1.

i = 3:

j = 0: f(3, 0) = f(2, 1) = 0
j = 1: f(3, 1) = f(2, f(3, 0)) = f(2, 0) = 3
j = 2: f(3, 2) = f(2, f(3, 1)) = f(2, 3) = 4
j = 3: f(3, 3) = f(2, f(3, 2)) = f(2, 4) = 1
j = 4: f(3, 4) = f(2, f(3, 3)) = f(2, 1) = 0

Third row: 0, 3, 4, 1, 0.

i = 4:

j = 0: f(4, 0) = f(3, 1) = 3
j = 1: f(4, 1) = f(3, f(4, 0)) = f(3, 3) = 1
j = 2: f(4, 2) = f(3, f(4, 1)) = f(3, 1) = 3
j = 3: f(4, 3) = f(3, f(4, 2)) = f(3, 3) = 1
j = 4: f(4, 4) = f(3, f(4, 3)) = f(3, 1) = 3

Fourth row: 3, 1, 3, 1, 3.

Hmm, interesting. Let me see if there's a pattern here. Let's list the rows for i from 0 to 4:

i=0: 1, 2, 3, 4, 0

i=1: 2, 3, 4, 0, 1

i=2: 3, 0, 2, 4, 1

i=3: 0, 3, 4, 1, 0

i=4: 3, 1, 3, 1, 3

Looking at these, maybe the function cycles with some period. Let's see.

Looking at the first few rows:

i=0: 1,2,3,4,0

i=1: 2,3,4,0,1

i=2: 3,0,2,4,1

i=3: 0,3,4,1,0

i=4: 3,1,3,1,3

Hmm, is there a cycle here? Let's check if i=4 is similar to any previous i.

Looking at i=4: 3,1,3,1,3. That's not the same as i=0,1,2,3. Maybe the cycle is longer.

Let me compute i=5:

j=0: f(5, 0) = f(4, 1) = 1

j=1: f(5,1) = f(4, f(5,0)) = f(4,1) = 1

j=2: f(5,2) = f(4, f(5,1)) = f(4,1) = 1

j=3: f(5,3) = f(4, f(5,2)) = f(4,1) = 1

j=4: f(5,4) = f(4, f(5,3)) = f(4,1) = 1

So i=5: 1,1,1,1,1

That's interesting. All ones. Let's check i=6:

j=0: f(6,0) = f(5,1) = 1

j=1: f(6,1) = f(5, f(6,0)) = f(5,1) = 1

Similarly, all j from 0 to 4 will result in 1, so i=6: 1,1,1,1,1.

So from i=5 onwards, it seems that all f(i,j) = 1. Wait, but let's check that again.

Wait, when i=5, j=0: f(5,0) = f(4,1) = 1

Then for j=1: f(5,1) = f(4, f(5,0)) = f(4,1) = 1

Similarly, for j=2: f(5,2) = f(4, f(5,1)) = f(4,1) = 1

Same for j=3 and 4. So all f(5,j) = 1.

Then for i=6:

j=0: f(6,0) = f(5,1) = 1

j=1: f(6,1) = f(5, f(6,0)) = f(5,1) = 1

Same for j=2,3,4. So f(6,j) =1 for all j.

Similarly, i=7: same as i=6. So from i=5 onwards, all rows are [1,1,1,1,1].

Wait, but that seems too straightforward. Let me check if this is correct. Let's go back to i=4, j=2: f(4,2) = f(3, f(4,1)) = f(3,1) = 3. Then i=5, j=0: f(5,0)=f(4,1)=1. Then i=5, j=1: f(5,1)=f(4,1)=1. Then for i=5, j=2: f(5,2)=f(4,1)=1. Since all f(5,j) are 1, then for i=6, j=1: f(6,1)=f(5,1)=1. Similarly, all j's in i=6 would be 1. So yes, that seems correct. So once we hit i=5, all subsequent i's have f(i,j)=1 for all j.

So if that's the case, then for any i >=5, f(i,j)=1. Therefore, f(2015,2)=1. But wait, let me confirm this with i=5 and i=6.

Wait, for i=5, j=0: f(5,0)=f(4,1)=1. Then for j=1: f(5,1)=f(4,1)=1. For j=2: f(4,1)=1. Similarly, all j's in i=5 result in 1. Then for i=6, j=1: f(6,1)=f(5,1)=1. So all j's in i=6 are 1. So yes, once we get to i=5, all the subsequent i's (i=5,6,7,...) have all f(i,j)=1.

Therefore, regardless of how large i is, as long as i >=5, f(i,j)=1. Therefore, f(2015,2)=1.

But let me check if this is consistent with the initial computations. Wait, when i=5, all j's are 1. Then for i=6, same thing. So f(2015,2)=1. That seems to be the answer. But let me check if there's a mistake here. Let me verify with i=5 and j=4.

Wait, f(5,4) is f(4, f(5,3)). f(5,3) is f(4, f(5,2)) = f(4,1) =1. Then f(5,4) is f(4,1)=1. So yes, all f(5,j)=1. Then for i=6, j=0: f(6,0)=f(5,1)=1. Then j=1: f(6,1)=f(5,1)=1. Similarly, all j's in i=6 are 1. So this seems to hold.

Therefore, the answer should be 1. However, let me check if the function is indeed reaching 1 at i=5. Let's compute f(4,1) again. For i=4, j=1: f(4,1)=f(3, f(4,0))=f(3,3)=1. Then f(5,0)=f(4,1)=1. Then f(5,1)=f(4,1)=1, etc. So yes, that's correct.

Therefore, the answer is 1. But let me check if the problem states that i is a positive integer, and j is between 0 and 4. So yes, 2015 is a positive integer, and 2 is between 0 and 4. Therefore, the answer is 1.

Wait, but let me check if there's any mistake in the recursive steps. Let's take i=5, j=2: f(5,2)=f(4, f(5,1))=f(4,1)=1. Then for i=5, j=3: f(5,3)=f(4, f(5,2))=f(4,1)=1. Similarly, j=4: f(5,4)=f(4,1)=1. So all j=0 to 4 in i=5 are 1. Then i=6: j=0: f(6,0)=f(5,1)=1. j=1: f(6,1)=f(5,1)=1. So yes, 1 for all j.

Therefore, the answer is 1. So f(2015,2)=1. Therefore, the final answer is \boxed{1}.

**Final Answer**
\boxed{1}
Therefore, the final answer is $\boxed{1}$.
</think>