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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, it is left we only have to consider case 2) only. 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.
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 between [0..in the first 2^28] terms is a(106956964) = 369^2 (= 136161). Notably, this is "relatively close" to n, and This further encourages one to believe that Goldbach's binary conjecture holds true (but this doesn't explicitly demonstrate new knowledge). - Daniel Mikhail, Nov 23 2020
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(PARI) a(n) = {my(lim = if (issquare(n), sqrtint(n)-1, sqrtint(n))); for (k=0, lim, if (bigomega(n-k^2) == 2, return (k^2)); ); return (2); } \\ Michel Marcus, Nov 26 2020
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The largest a(n) term found between [0..2^28] is a(106956964)=369^2 (136161). Notably, this is "relatively close" to n, and further encourages one to believe that Goldbach's binary conjecture holds true (but this doesn't explicitly demonstrate new knowledge). - Daniel Mikhail, Nov 23 2020
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Daniel Mikhail, <a href="https://raw.githubusercontent.com/mikhaidn/SemiprimeCalculations/main/Summary∆k^2, away from a semiprime for all ∆k's found between [05..2^28]</a>
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