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List of the first fifty prime gaps:

1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4.

My conjecture is that the sum of consecutive prime gaps is always prime whenever a prime gap of 2 is added.

$$ 1 + 2 = 3 $$ $$ 1 + 2 + 2 = 5 $$ $$ 1 + 2 + 2 + 4 + 2 = 11 $$ $$ 1 + 2 + 2 + 4 + 2 + 4 + 2 = 17 $$ $$ 1 + 2 + 2 + 4 + 2 + 4 + 2 + 4 + 6 + 2 = 29 $$

I don't know if this is meaningful or how to go about testing it completely (I've tested it up to 461) so I'll just leave this here and see what comes of it.

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    $\begingroup$ It is just the definition of prime gap. There is no gap leading up to 2, so adding 2 to the sum of the gaps will give you the next prime. $\endgroup$
    – gamma
    Commented Aug 14, 2014 at 1:11
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    $\begingroup$ Sounds like something for the OEIS. (I looked it up, got no results). $\endgroup$
    – user153918
    Commented Aug 14, 2014 at 15:23
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    $\begingroup$ To be more precise, I did get a red herring result in my search: oeis.org/A040998 $\endgroup$
    – user153918
    Commented Aug 14, 2014 at 15:40
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    $\begingroup$ @AlonsodelArte Question was updated. The sequence is A040998. $\endgroup$
    – Jeppe Stig Nielsen
    Commented Sep 6, 2014 at 6:28
  • $\begingroup$ @AlonsodelArte No? it's just the lesser twin primes. oeis.org/A001359 $\endgroup$
    – Tony
    Commented Sep 19, 2014 at 4:35

3 Answers 3

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Set $g_n=p_{n+1}-p_n$, where $p_n$ is the series of prime numbers, with $p_1=2$.
Then $$ p_1+\sum_{i=1}^n g_i=\sum_{i=1}^n g_i+2=p_{n+1}. $$ So the conjecture is obviously true, but not useful.

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Let $p_n$ denote the $n^{\text{th}}$ prime. Then $p_{n+1}-p_n$ is the $n^{\text{th}}$ prime gap. The sum of the first $k$ prime gaps is $$\sum_{n=1}^k(p_{n+1}-p_n) = p_{k+1} - p_1 = p_{k+1} - 2.$$ Now, if the $k^{\text{th}}$ prime gap is $2$, that is $p_{k+1} - p_k = 2$, then $p_{k+1} - 2 = (p_k + 2) - 2 = p_k$ which is prime as you've noticed.

A pair of consecutive primes $p_n$, $p_{n+1}$ which differ by two (i.e. the $n^{\text{th}}$ prime gap is two) are said to be twin primes. It is still unknown as to whether there are infinitely map twin primes.

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The first gap is 3 - 2 = 1. The sum of gaps up to the first gap is 1, that is 2 less than 3.

The second gap is 5 - 3 = 2. The sum of gaps up to the second gap is 3, that is 2 less than 5.

The third gap is 7 - 5 = 2. The sum of gaps up to the third gap is 5, that is 2 less than 7.

The fourth gap is 11 - 7 = 4. The sum of gaps up to the fourth gap is 9, that is 2 less than 11.

You should be able to see a pattern now: The sum of the first n gaps is always two less than the n'th prime. And that's due to the way the gaps are calculated: The (n+1)st gap is the difference between the n+1st and the nth prime; adding this to the sum so far which is two less than the nth prime will give a result that is two less than the nth prime.

And of course every time the gap is two, two less than the nth prime happens to be the n-1st prime, which explains everything nicely.

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