%I #9 Jul 12 2014 22:44:20
%S 2,1,1,2,1,2,2,1,2,2,1,2,2,1,3,3,2,1,3,2,3,3,2,3,2,2,3,2,2,3,3,2,2,3,
%T 2,3,3,2,2,4,2,4,2,2,4,2,2,1,4,2,4,4,2,4,4,2,4,2,2,1,2,1,1,4
%N Largest k such that the smallest prime satisfying Goldbach's conjecture is less than or equal to (2n)^(1/k).
%C The 1's appear as in A244408.
%H Jens Kruse Andersen, <a href="/A245077/b245077.txt">Table of n, a(n) for n = 2..10000</a>
%e For n=5 we have 3+7=10. As rt3(10)<3<sqrt(10), a(5)=2.
%o (PARI) for (n=2, 100, p=2; while(!isprime(2*n-p), p=nextprime(p+1)); k=1; while(p<=(2*n)^(1/k), k++); print1(k-1", ")) \\ _Jens Kruse Andersen_, Jul 12 2014
%Y Cf. A244408.
%Y Cf. A020481, A002373, A025018, A025019.
%K nonn
%O 2,1
%A _Jon Perry_, Jul 11 2014
%E Definition corrected by _Jens Kruse Andersen_, Jul 12 2014
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