OFFSET
1,1
COMMENTS
This sequence is related to a stronger form of Goldbach conjecture, and the formulation of the conjecture is the following.
Conjecture: Let
Gs(m) be the number of unordered pairs of odd prime numbers p and q such that an even number 2m can be written as the sum of p and q, or, Gs(m) = # { (p,q) | 2m = p+q }, where p <= q;
d be half of the minimum difference between q and p, or, d = min((q-p)/2);
r be the ratio of d/m, or, r = d/m; and
a(n) be the n-th number such that r = d/m is decreasing, or, r(m) < r(a(n+1)), if a(n) <= m < a(n+1), where n = 0, 1, 2, 3, .... and a(0)=3.
The conjecture states that
Gs(m) >= 1 if r(m) >= r(a(n+1)) where a(n) <= m < a(n+1).
EXAMPLE
For even numbers 2m >= 6 (6 is the smallest even number that can be written as the sum of two odd primes), the list of m is:
m = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, ...}.
The corresponding values of d and r, according to the definition, are given in the following two lists:
d = {0, 1, 0, 1, 0, 3, 2, 3, 0, 1, 0, 3, 2, 3, 0, 1, 0, 3, 2, 9, 0, 5, 6, 3, 4, 9, 0, 1, 0, 9, 4, 3, 6, 5, 0, 9, 2, 3, 0, 1, 0, 3, 2, 15, ...}, and
r = {0, 1/4, 0, 1/6, 0, 3/8, 2/9, 3/10, 0, 1/12, 0, 3/14, 2/15, 3/16, 0, 1/18, 0, 3/20, 2/21, 9/22, 0, 5/24, 6/25, 3/26, 4/27, 9/28, 0, 1/30, 0, 9/32, 4/33, 3/34, 6/35, 5/36, 0, 9/38, 2/39, 3/40, 0, 1/42, 0, 3/44, 2/45, 15/46, ...}.
In the list of r, the first number that is bigger than all the preceding numbers is r = 9/22, which is corresponding to the number m = 22 in the list of m. Therefore, the first number of the sequence is 22, or a(1) = 22.
In the range of (9/22, 15/46], r= 15/46 is the biggest number. Since r = 15/46 corresponds to m = 46, the 2nd number of the sequence is 46, or a(2) = 46.
The first number in the list of m, 3, is defined as the zeroth term of the sequence, or a(0) = 3.
PROG
(PARI) mindiff(n) = {forstep(k=n/2, 1, -1, if (isprime(k) && isprime(n-k), return(n-2*k)); ); }
upto(n) = res=List(); r=0; forstep(i=n, 1, -1, c= mindiff(2*i) / (2*i); if(c>r, r=c; listput(res, i))); Vecrev(Vec(res)) \\ David A. Corneth, Jun 02 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Ya-Ping Lu, May 30 2020
STATUS
approved