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Old 2019-03-07, 18:14   #8
Dr Sardonicus
 
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Feb 2017
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Obviously, the gaps, being symmetrical around N, will be the same on both sides of N as you move away from N. Equally obviously, if 3|N, the first gaps p-N and N-p are not divisible by 3. That'll teach me to post before my blood caffeine is up to working levels. At least until the next time I forget and do it again.

But the observation of two possible sequences of non-zero residues is right. And now, with sufficient caffeine in my system, I can try to figure out how many sequences of mod 3 residues are possible in a sequence of k gaps on one side of N, assuming 3|N. The first gap has to have a non-zero residue (mod 3). Say it's 1. Each of the other k-1 gaps can have either a zero residue, or a non-zero residue. The succeeding non-zero residues have to be from the repeating sequence 1,2,1,2,... so, having picked j of the remaining k-1 slots, which can be done in binomial (k-1,j) ways, there is only one way to fill them in with non-zero residues. Thus, having chosen the residue 1 for the first gap, there are 2^(k-1) ways to assign a sequence of k residues (mod 3). If we choose the initial residue 2 instead, we get another 2^(k-1) possibilities. So, 2^k in all.

Now, we look at the 2*k - 1 gaps between a sequence of 2*k consecutive primes. We'll assume we start with a prime p > 3. (In the case of 2*k primes symmetrically placed about N, the two gaps between N and the nearest primes are joined into one gap between consecutive primes.) There are 2 possible residues (mod 3) for the first gap. Choosing one, the remaining gaps (mod 3) can be filled in either with 0's, or the initial terms of whichever sequence of nonzero residues 1,2,1,2... or 2,1,2,1... avoids multiples of 3. By the same argument as before, the result is a total of 2^(2*k-1) possible sequences of residues (mod 3).

So, if 3|N, the symmetry condition of k primes either side of N restricts the sequences of residues (mod 3) to 2^k out of 2^(2*k-1) possibilities for a sequence of 2*k consecutive primes. If N is not divisible by 3 (and the smallest prime is greater than 3), the symmetry condition restricts consideration to one sequence (all zeroes) out of 2^(2*k-1) possible sequences of residues (mod 3) of the sequence of 2*k - 1 gaps between 2*k consecutive primes.

I have no clue whether the possible sequences of residues (mod 3) of of gaps between a given number of consecutive primes are in any sense "equally distributed."
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