what is Prime Gaps relationship with number 6? Out of the 78499 prime number under 1 million. There are 32821 prime gaps (difference between two consecutive prime numbers) of a multiple 6. A bar chart of differences and frequency of occurrence shows a local maximum at each multiple of 6. Why is 6 so special? 

 A: All prime numbers except 2 and 3 are of the form $6k±1$, so whenever you fall on a pair $6k+1$, $6l+1$ their difference will be a multiple of $6$, same goes for a pair $6k-1$,$6l-1$.
http://en.wikipedia.org/wiki/Primality_test#Naive_methods
A: Excepting the first two gaps, all prime gaps are between numbers that are either $1$ or $5$ modulo $6$. Under the assumption that both cases are equally likely, half the prime gaps will be between numbers in the same class, and therefore of size $0$ modulo $6$, and the other half will be between numbers in different classes, which split up into sizes that are $2$ and $4$ modulo $6$. Since each of the latter cases only gets one quarter of the total, it is clear that ignoring all other factors, gaps that are $2$ or $4$ modulo $6$ are about half as likely to occur as gaps of the same approximate magnitude that are $0$ modulo $6$. You can check this in your chart. (Particular gap sizes are also subject to influences of other primes than $2$ or $3$, which explains some other irregularities one can observe.)
A: To provide a different perspective on Vhailor's answer: note that if $p$ is a prime $\gt 3$, then $p+6k$ is guaranteed not to be divisible by $2$ or $3$ for any $k$; in effect these gaps are 'pre-sieved' to weed out possible multiples of $2$ and $3$ that could keep the number at the other end from being prime.  If you expanded your chart out further you would see similar spikes at the multiples of $30$, since those numbers are also 'pre-sieved' for $5$.  (In fact, if you were to expand your table out to all the prims less than $2\times 10^{35}$, you would find the total number of gaps of length $30$ to be more than the number of gaps of length $6$ - see http://mac6.ma.psu.edu/primes/ for the details!)
A: 6 is "special" because it is 3#  (primorial, products of  the first n primes) = 2*3.  The first primorial is 2, and generates all even and odd numbers through 2x + (0,1).  The second primorial, 6, generates all primes greater than 3 through 6x + (1,5).  Disclaimer: this formula also generates composites.  
The third primorial generates all primes greater than 5 with 30x + (1,7,11,13,17,19,23,29)... And so on
The high concentration of prime distances equal to six has more to do with the rarity of "new composites" eliminated by large prime seives.  997, the largest prime sieve to establish all primes under a million, only removes 1 composite under a million.  As you get further from 0, the prime patterns are largely preserved.  This has led to the twin prime and k-tuple conjectures, among others.
A: Here is a recent paper on how the prime numbers are a perfect example of harmonic acceleration between two poles, where the multiples of 6 (6z, acceleration) is the second derivative of z^3. 
http://www.wseas.us/e-library/conferences/2012/CambridgeUSA/MATHCC/MATHCC-05.pdf
Yes, I am the author.
A: Gaps of 6 are constructed from gaps of 2 and 4.  Gaps of 2 are the result of sieving out the even numbers and even numbers account for all gaps of 2.  The gaps of 4 are the result of sieving out all multiplies of 3 and no other numbers creates gaps of 4.  The prime number number 5 is the first number to construct gaps of 6 other prime numbers also construct gaps of 6 resulting in the large spike in the number of gaps of 6.  Gaps of 2 and 4 become almost evenly distributed in the list of numbers.  Many of the gaps of 4 will have one or two twin primes next to it.  The total number of  gaps is 1 less than the total number of prime numbers.     
A: take any integer $n> 3$, and divide it by $6$.  That is, write
$n = 6q + r$
where $q$ is a non-negative integer and the remainder $r$ is one of $0$, $1$, $2$, $3$, $4$, or $5$.
If the remainder is $0$, $2$ or $4$, then the number $n$ is divisible by $2$, and can not be prime.
If the remainder is $3$, then the number $n$ is divisible by $3$, and can not be prime.
So if $n$ is prime, then the remainder $r$ is either


*

*$1$ (and $n = 6q + 1$   is one more than a multiple of six), or

*$5$ (and $n = 6q + 5 = 6(q+1) - 1$   is one less than a multiple of six).

