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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 February 20 00:28 EST 2019. Contains 320329 sequences. (Running on oeis4.)