What you have is what's known in some circles as a linear recurrence. That is, you have a sequence where the general term can be expressed as an appropriate combination of previous terms.

The simplest way of going about it, as has been previously noted, is to use either of Nest[]/NestList[], like so:

NestList[Append[#, Total[Take[#, -2]]] &, {0, 2, 2, 4}, 6] {{0, 2, 2, 4}, {0, 2, 2, 4, 6}, {0, 2, 2, 4, 6, 10}, {0, 2, 2, 4, 6, 10, 16}, {0, 2, 2, 4, 6, 10, 16, 26}, {0, 2, 2, 4, 6, 10, 16, 26, 42}, {0, 2, 2, 4, 6, 10, 16, 26, 42, 68}} 

where Take[list, -2], takes the last two (hence the negative sign) elements, and Total[list] sums the two elements up.

However, it should be said that there is a function called, appropriately enough, LinearRecurrence[], that does this recursion in an efficient manner. Here is how to use it:

LinearRecurrence[{1, 1}, {0, 2}, 10] {0, 2, 2, 4, 6, 10, 16, 26, 42, 68} 

To explain the notation a bit, LinearRecurrence[{p, q}, {a, b}, k] generates k terms, starting with the first two terms a and b, and then successively generates a + b, a + 2 b, 2 a + 3 b, .... You can see this by feeding symbolic arguments:

LinearRecurrence[{1, 1}, {a, b}, 10] {a, b, a + b, a + 2 b, 2 a + 3 b, 3 a + 5 b, 5 a + 8 b, 8 a + 13 b, 13 a + 21 b, 21 a + 34 b} 

In fact, your sequence can be expressed in terms of the famous Fibonacci numbers, which satisfy the same recursion, but with different starting points. I'll skip the (deep!) math, but here's a demonstration:

Table[2 (Fibonacci[n - 3] + Fibonacci[n - 2]), {n, 9}] {0, 2, 2, 4, 6, 10, 16, 26, 42, 68} 

What you have is what's known in some circles as a linear recurrence. That is, you have a sequence where the general term can be expressed as an appropriate combination of previous terms.

The simplest way of going about it, as has been previously noted, is to use either of Nest[]/NestList[], like so:

NestList[Append[#, Total[Take[#, -2]]] &, {0, 2, 2, 4}, 6] {{0, 2, 2, 4}, {0, 2, 2, 4, 6}, {0, 2, 2, 4, 6, 10}, {0, 2, 2, 4, 6, 10, 16}, {0, 2, 2, 4, 6, 10, 16, 26}, {0, 2, 2, 4, 6, 10, 16, 26, 42}, {0, 2, 2, 4, 6, 10, 16, 26, 42, 68}} 

where Take[list, -2], takes the last two (hence the negative sign) elements, and Total[list] sums the two elements up.

However, it should be said that there is a function called, appropriately enough, LinearRecurrence[], that does this recursion in an efficient manner. Here is how to use it:

LinearRecurrence[{1, 1}, {0, 2}, 10] {0, 2, 2, 4, 6, 10, 16, 26, 42, 68} 

To explain the notation a bit, LinearRecurrence[{1, 1}, {a, b}, k] generates k terms, starting with the first two terms a and b, and then successively generates a + b, a + 2 b, 2 a + 3 b, .... You can see this by feeding symbolic arguments:

LinearRecurrence[{1, 1}, {a, b}, 10] {a, b, a + b, a + 2 b, 2 a + 3 b, 3 a + 5 b, 5 a + 8 b, 8 a + 13 b, 13 a + 21 b, 21 a + 34 b} 

In fact, your sequence can be expressed in terms of the famous Fibonacci numbers, which satisfy the same recursion, but with different starting points. I'll skip the (deep!) math, but here's a demonstration:

Table[2 (Fibonacci[n - 3] + Fibonacci[n - 2]), {n, 9}] {0, 2, 2, 4, 6, 10, 16, 26, 42, 68}