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 A241922 Smallest k^2>=0 such that n-k^2 is semiprime, or a(n)=2 if there is no such k^2. 5
 2, 2, 2, 0, 1, 0, 1, 4, 0, 0, 1, 2, 4, 0, 0, 1, 2, 4, 4, 16, 0, 0, 1, 9, 0, 0, 1, 2, 4, 4, 9, 2, 0, 0, 0, 1, 4, 0, 0, 1, 16, 4, 4, 9, 36, 0, 1, 9, 0, 1, 0, 1, 4, 16, 0, 1, 0, 0, 1, 9, 4, 0, 1, 9, 0, 1, 9, 64, 0, 1, 9, 2, 4, 0, 1, 25, 0, 1, 64, 25, 4, 0, 1, 49, 0, 0, 0, 1, 4, 4, 0, 1, 0, 0, 0, 1, 4, 4, 4, 9, 16 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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 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. 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. LINKS Peter J. C. Moses, Table of n, a(n) for n = 1..1000 FORMULA a(A001358(n)) =0. CROSSREFS Cf. A000290, A001358, A100570, A152522, A152451, A156537. Sequence in context: A108867 A218604 A095767 * A071446 A071469 A071471 Adjacent sequences:  A241919 A241920 A241921 * A241923 A241924 A241925 KEYWORD nonn AUTHOR Vladimir Shevelev, May 01 2014 STATUS approved

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Last modified July 6 06:00 EDT 2020. Contains 335476 sequences. (Running on oeis4.)