cp's OEIS Frontend

Powers of 2 are not expressible as sums of two primes from this sequence. This is attained by a more economical algorithm than that for construction of A152451. If A(x) is the counting function for the terms a(n) <= x, then A(x) = pi(x) - O(x/(log^2(x)). It is known that the approximation of pi(x) by x/log(x) gives the remainder term as, at best, O(x/log^2(x)). Therefore beginning our process from m >= M (with arbitrarily large M), we obtain a sequence which essentially is indistinguishable from the sequence of all odd primes with the help of the approximation of pi(x) by x/log(x). Hence it is in principle impossible to prove the binary Goldbach conjecture by such an approximation of pi(x).

If n = m^2, m>=2, then the condition {a(n) differs from 2} is equivalent to the Goldbach binary conjecture. Indeed, if m^2 - k^2 is semiprime, then (m-k)*(m+k) = p*q, where p<=q are primes. Here we consider two possible cases. 1) m-k=1, m+k=p*q and 2) m-k=p, m+k=q. But in the first case k=m-1>m-p, i.e., more than k in the second case. In view of the minimality k, we only have to consider case 2). In this case we have m-/+k both are primes p<=q (with equality in case k=0) and thus 2*m = p + q. Conversely, let the Goldbach conjecture be true. Then for a perfect square n>=4, we have 2*sqrt(n)=p+q (p<=q are both primes). Thus n=((p+q)/2)^2 and n-((p-q)/2)^2=p*q is semiprime. Hence a(n) is a square not exceeding ((p-q)/2)^2.

Note that a(n)=2 for 1,2,3,12,17,28,32,72,...

All these numbers are in A100570. Thus the Goldbach binary conjecture is true if and only if A100570 does not contain perfect squares.

The largest term found in the first 2^28 terms is a(106956964) = 369^2 = 136161. This further encourages one to believe that Goldbach's binary conjecture holds true. - Daniel Mikhail, Nov 23 2020

A semiprime in Fermi-Dirac arithmetic is a product of two distinct terms of A050376, or, equivalently, an infinitary semiprime. The conjecture that every even number>=4 is a sum of two A050376 terms is a weaker form of the Goldbach conjecture; as such, it is natural to refer to it as a Goldbach conjecture in Fermi-Dirac arithmetic (FDGC).

Let us prove that the condition {a(m^2) differs from 2} is equivalent to the FDGC.

Indeed, from the FDGC for a perfect square n>=4, we have 2*sqrt(n)=p+q (pA050376 terms of the same parity). Thus n=((p+q)/2)^2 and n-((p-q)/2)^2=p*q is Fermi-Dirac semiprime. Hence, a(n)>=1 is a square not exceeding ((p-q)/2)^2. Thus the condition {a(m^2) differs from 2} is necessary for the truth of the FDGC.

Let us prove that the condition {a(m^2) differs from 2} is also sufficient. Indeed, a(m^2)-k^2 = p*q, where, say, pA050376, and p,q are of the same parity. If p,q are primes, then the proof repeats one in A241922. Let, e.g., p=s^2A050376). Consider two principal cases: 1) m-k = s, m+k = s*q; 2) m-k = s^2, m+k = q. In 1) k=m-s, in 2) k=m-s^2. In view of the minimality of k, we should accept 2) and thus m-k=p, m+k=q. So, 2*m=p+q as the FDGC requires.

The sequence of numbers n for which a(n)=2 begins 1, 2, 3, 4, 5, 6, 8, 20, ... (A241947).

The restriction a(n) <= n^2 + 1 allows one to make the sequence computable for n >= 1 and, at the same time, to somewhat agree with heuristic arguments for large n.

The sequence is based on a conjecture stronger than Goldbach's Conjecture: for arbitrarily large N there exists a number m(N) such that, for k > m(N), the number of unordered Goldbach representations (A002375) of 2*k is greater than N.

Heuristic arguments would imply that m(N) ~ N*log^2(2*N). Then, conjecturally, for n >= 3, a(n) < n*log^2(2*n).

The existence of a(n) for arbitrary n says that, if we remove from the sequence of primes an arbitrarily large number M of the first terms, the Goldbach representations remain for all sufficiently large even numbers.