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 A241927 Smallest k^2>=1 such that n-k^2 is semiprime p*q in Fermi-Dirac arithmetic (A176525) with additional requirement that, if n is a square, then p and q are of the same parity; or a(n)=2 if there is no such k^2. 4
 2, 2, 2, 2, 2, 2, 1, 2, 1, 4, 1, 4, 1, 4, 1, 1, 9, 4, 1, 2, 1, 1, 1, 4, 4, 4, 1, 1, 1, 4, 4, 4, 1, 1, 1, 1, 1, 4, 1, 1, 9, 4, 4, 9, 1, 1, 1, 4, 4, 4, 1, 1, 1, 4, 4, 1, 9, 1, 1, 9, 4, 4, 1, 1, 1, 1, 4, 4, 1, 1, 9, 4, 4, 9, 1, 1, 1, 1, 4, 4, 4, 25, 1, 4, 9, 1, 1, 1, 4, 4, 4, 1, 1, 1, 1, 1, 4, 4, 1, 1, 1, 4, 4, 4, 25 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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 (p=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, p

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Last modified October 23 08:57 EDT 2019. Contains 328345 sequences. (Running on oeis4.)