

A241922


Smallest k^2>=0 such that nk^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
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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 (mk)*(m+k) = p*q, where p<=q are primes. Here we consider two possible cases. 1) mk=1, m+k=p*q and 2) mk=p, m+k=q. But in the first case k=m1>mp, 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((pq)/2)^2=p*q is semiprime. Hence, a(n) is a square not exceeding ((pq)/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



