cp's OEIS Frontend

A307211 a(n) = the "maximum first open number" for prime(n).

%I A307211 #19 Jul 01 2019 02:01:39
%S A307211 2,6,12,24,42,75,90,150,180,216,312,339,447,519,615,660,783
%N A307211 a(n) = the "maximum first open number" for prime(n).
%C A307211 This sequence is related to Goldbach's Conjecture.
%C A307211 Definition. Let P = prime(n). Choose one remainder and its negative for each prime up to and including P. Sieve the numbers 1, 2, 3, 4, ... to remove numbers congruent to either chosen remainder modulo the respective prime. Call the numbers left "open", and the smallest open number the "first open number" for that P and that choice of remainders. The maximum first open number for P is the largest first open number for any such choice of remainders.
%C A307211 Equivalently, let P = prime(n). Any positive integer c has a set of remainders modulo each prime up to and including P. Call a positive integer d "open" for c and P, if d has no remainder which is the same as, or is the negative of, any remainder of c modulo any prime up to and including P. There is a smallest, or first, open number d for any c. The maximum first open number for P is the largest first open number d for any positive integer c. (Only numbers c from 1 to P#, the product of the primes up to and including P, have to be considered.)
%C A307211 A problem is to find an upper limit for a(n) in terms of prime(n).
%C A307211 The ratios a(n)/prime(n), n = 1 to 14, are 1, 2, 2.4, 3.4, 3.8, 5.8, 5.3, 7.9, 7.8, 7.4, 10.1, 9.2, 10.9, 12.1, so a(n) appears to grow faster than prime(n).
%C A307211 The ratios log(a(n))/log(prime(n)), n = 1 to 14, are 1, 1.63, 1.54, 1.63, 1.56, 1.68, 1.59, 1.70, 1.66, 1.60, 1.67, 1.61, 1.64, 1.66, which appear to be bounded.
%C A307211 Conjecture 1: a(n) <= prime(n)^1.75.
%C A307211 Conjecture 2: a(n) <= prime(n) * (prime(n) - 1) / 2, n >= 5.
%C A307211 More terms are needed to check whether these conjectures are true for larger n.
%C A307211 Either conjecture implies that Goldbach's conjecture is true. (Stated as every even number greater than 6 is the sum of two different primes where 1 is not prime.)
%C A307211 Consider the sum (c - d) + (c + d) = 2c, where d is open for c and prime P, the smallest prime such that the primes up to and including P are sufficient to test whether two numbers that add to 2c are prime. If d were small enough, c - d is positive and both c - d and c + d would be prime, because c and d and also -c and d have no common remainders modulo every prime up to and including P.
%C A307211 Either conjecture would assure the existence of an open number d which is small enough, for every sufficiently large c.
%C A307211 For example, for P large enough, P^1.75 <= (P^2 - 3)/2. For a particular c, choose P so it is the largest prime with P <= sqrt(2c - 3). Rearranged, this gives (P^2 - 3)/2 <= c - 3. Thus d <= P^1.75 <= (P^2 - 3)/2 <= c - 3, so that c - d >= 3 is positive. Therefore Goldbach's Conjecture follows from Conjecture 1.
%C A307211 I think that conjectures 1 and 2 would imply that, for every gap 2n, there are an infinite number of prime pairs with that gap. I think they would also give an upper bound for the gap between pairs of primes with gap 2n.
%H A307211 Sally M. Moite, <a href="http://vixra.org/abs/1906.0282">Maximum First Open Numbers and Goldbach's Conjecture</a>, viXra.org:1906.0282 (2019). Additional comments. Unedited.
%H A307211 <a href="/index/Go#Goldbach">Index entries for sequences related to Goldbach conjecture</a>
%F A307211 a(n) = max for all {r(1), ..., r(n)} min d = d({r}) where d >= 1, d !== r(i) (mod prime(i)), d !== -r(i) (mod prime(i)), i = 1, ..., n.
%F A307211 a(n) = max for all c >= 1, min d = d(c) where d >= 1, d !== c (mod p) and d !== -c (mod p) for all p, p prime, p <= prime(n).
%e A307211 Let n = 4, so P = 7.
%e A307211 Choose, for example, remainders 1 (mod 2), 0 (mod 3), +-1 (mod 5), +-2 (mod 7).
%e A307211 Remove odd numbers and numbers divisible by 3 from 1, 2, 3,..., 49 (which should be enough numbers to sieve according to the conjectures) leaving 2, 4, 8, 10, 14, 16, 20, 22, 26, 28, 32, 34, 38, 40, 44, 46.
%e A307211 Then remove numbers congruent to +-1 (mod 5), which leaves 2, 8, 10, 20, 22, 28, 32, 38, 40.
%e A307211 Finally remove numbers congruent to +-2 (mod 7), which leaves the "open" numbers 8, 10, 20, 22, 28, 32, 38, 46. The "first open number" is 8.
%e A307211 There are 2 * 2 * 3 * 4 = 48 ways of choosing remainders for P = 7 (0 or 1 for 2, 0 or +-1 for 3, 0, +-1 or +-2 for 5, 0, +-1, +-2 or +-3 for 7).
%e A307211 The maximum first open number for 7 is 24, for remainders 1 (mod 2), +-1 (mod 3), +-2 (mod 5) and +-1 (mod 7).
%e A307211 For another example, let n = 3, so P = 5. For numbers c, one need only consider the numbers 1 to 30 to account for all possible combinations of remainders mod 2, 3, and 5. The first open numbers for each of these numbers, for P = 5, are 12, 9, 4, 3, 6, 5, 6, 9, 2, 3, 12, 1, 6, 3, 2, 3, 6, 1, 12, 3, 2, 9, 6, 5, 6, 3, 4, 9, 12, 1 respectively. Therefore, for n = 3, the "maximum first open number" a(3) is 12.
%o A307211 (Outline)
%o A307211 a(n)=1
%o A307211 P=prime(n)
%o A307211 For each permutation m(n) of 1 to n
%o A307211   i()={1,2,3,...,P^2}
%o A307211   for j=1,n
%o A307211     r=i(1) mod p=prime(m(j))
%o A307211     eliminate numbers congruent to r or -r mod p from i()
%o A307211   next j
%o A307211   if i(1) > a(n)
%o A307211   a(n)=i(1)
%o A307211 next permutation
%Y A307211 Cf. A000040, A002110, A005867.
%K A307211 nonn,more
%O A307211 1,1
%A A307211 _Sally Myers Moite_, Mar 28 2019
%E A307211 a(15)-a(17) from _Bert Dobbelaere_, Jun 02 2019