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$$q = x_a({p_a}^2)+4$$

Let $p_n$ be a sequence of consecutive prime numbers starting from 3. ($p_a$ represents a prime number in the sequence, and $x_a$ is the corresponding $x$ to the prime number.)

We are able to set nonzero natural number $x_a$ freely. How does one find such nonzero natural number $q$ that satisfies the aforementioned given the number of prime numbers in the sequence?

Also, what happens if

1) we allow $x_a$ and $q$ to be nonzero integer, not just natural number?

2) $p_n$ is defined as a sequence of consecutive ${p_a}^{b_a}$ where $p_a$ is defined as aforementioned We allow to set nonzero natural number $b_a$ freely. In this the equation changes to $q = x_a({p_a}^{2b_a})+4$.

3) If we remove "consecutive" from the definition of $p_n$ (either the first definition or the definition in 2).)

4) Combination of 1) and 2) or combination of 1) and 3)

Add: Is there any such $q$ that is smaller than the product of numbers $p_a$ in case of the origianl case? Also is there any such $q$ that is smaller than the product of numbers ${p_a}^{b_a}$ in case of 2)?

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    This doesn't make much sense. "Let $p_n$ be a sequence of consecutive prime numbers starting from 3." What do you mean, *a* sequence? There is only *one* sequence of consecutive prime numbers starting from 3, namely, the sequence 3, 5, 7, 11, 13, and so on.2012-11-08
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    I quoted exactly what you wrote.2012-11-08
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    I was just saying that $p_a$ is some prime number of the sequence. It's not $a$ sequence.2012-11-08
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    $p_a$ is a number in the sequence. $p_n$ is a sequence, in fact, *the* sequence of primes. Enough of this squabbling --- any comments/questions/whatever concerning the answer I posted?2012-11-09

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