There are multiple answers to this question because a finite number of terms do not precisely determine the sequence. For example, one could fit a polynomial f(x) of degree 10 to this sequence so that f(i) was equal to the ith term.

However here is another approach

First lets let [x] be the greatest integer in x so that, for example

[1/2] = 0, [5/2] = 2, and so on. Note that if k is an even number [k/2] = k/2, and if k is an odd number [k/2] = (k-1)/2 , and [0] = 0.

I will assume the sequence starts with S_{0},

then the m^{th} term in the sequence S_{m} is given by the formula

S_{m} = (2^{(m - 2[m/2])})∑_{[i≥0, i≤[m/2])}2^{i}

The way this formula works is that

(1) for even numbers m (including zero) the leading factor becomes 1, and for odd numbers it's 2.

(2) for even numbers the summation goes from 0 to m/2; for odd numbers it goes from 0 to (m-1)/2

So, for example, the 7th number in the sequence would be 2*(2^{3}+2^{2}+2^{1}+2^{0}) = 30;

and the 10th number in the sequence would be 1*(2^{5}+2^{4}+2^{3}+2^{2}+2^{1}+2^{0)} = 63

The sum of the powers of 2 from 0 to [m/2] can also be expressed as 2^{([m/2]+1)}-1, so we could rewrite the formula as

S_{m} = (2^{(m - 2[m/2])})*(2^{([m/2]+1)}-1)

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So the 2010^{th} term requested in the problem, since we defined the first term in the sequence to be S_{0}, would be S_{2009}

S_{2009} = 2*(2^{1005}-1) = 2^{1006}-2

I couldn't find any calculator that would compute this number without rounding. It's a number with about 303 digits.

## Comments

^{1006}= 2^{1004}*2^{4}, and 1004/4 = 251. This means that 2^{1006}ends in the digit 4, which means that the Wolfram answer ends with the correct digit.