Consecutive Prime Gap Sum (Amateur)

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.