Consider a hypothetical coin where the probability of getting the previous result, is half the probability of obtaining the previous result before it was tossed. For example, let \(p_{n}\) be the probability of getting heads in the \(n^\text{th}\) coin toss, and so \((1-p_{n})\) is the probability of getting tails in the \(n^\text{th}\) coin toss. Then,

- If the result of the \(n^\text{th}\) coin toss is heads then \(p_{n+1}=\dfrac{p_{n}}{2}\).
- If the result of the \(n^\text{th}\) coin toss is tails then \((1-p_{n+1})=\dfrac{1-p_{n}}{2}\) or \(p_{n+1}=\dfrac{1+p_{n}}{2}\).

The coin is taken and tossed until two consecutive heads are obtained. If \(E(HH)\) is the expected number of tosses to get two consecutive heads, find \(\left\lfloor 1000E(HH) \right\rfloor\).

**Details and Assumptions**:

For the first toss, the probability of getting heads and the probability of getting tails are both 0.5.

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