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A237720 Number of primes p <= (n+1)/2 with floor( sqrt(n-p) ) prime.

%I A237720 #16 Feb 12 2014 05:27:37
%S A237720 0,0,0,0,0,1,2,2,3,3,3,3,4,4,4,4,4,3,2,2,2,2,1,1,2,2,2,3,2,3,3,4,4,4,
%T A237720 4,5,5,5,4,4,3,4,3,4,4,4,3,4,3,3,4,5,4,5,4,5,6,6,5,6,7,8,8,8,7,7,5,6,
%U A237720 5,5
%N A237720 Number of primes p <= (n+1)/2 with floor( sqrt(n-p) ) prime.
%C A237720 Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 6, 23, 24, 111, 112, ..., 120.
%C A237720 (ii) For any integer n > 2, there is a prime p < n with floor(sqrt(n+p)) prime.
%C A237720 Note that floor(sqrt(n)) is the number of squares among 1, ..., n.
%C A237720 See also A237705, A237706 and A237721 for similar conjectures.
%H A237720 Zhi-Wei Sun, <a href="/A237720/b237720.txt">Table of n, a(n) for n = 1..10000</a>
%e A237720 a(6) = 1 since 2 and floor(sqrt(6-2)) = 2 are both prime.
%e A237720 a(23) = 1 since 11 and floor(sqrt(23-11)) = 3 are both prime.
%e A237720 a(24) = 1 since 11 and floor(sqrt(24-11)) = 3 are both prime.
%e A237720 a(27) = 2 since 2 and floor(sqrt(27-2)) = 5 are both prime, and 13 and floor(sqrt(27-13)) = 3 are both prime.
%e A237720 a(n) = 1 for n = 111, ..., 116 since 53 and floor(sqrt(n-53)) = 7 are both prime.
%e A237720 a(n) = 1 for n = 117, 118, 119, 120 since 59 and floor(sqrt(n-59)) = 7 are both prime.
%t A237720 q[n_]:=PrimeQ[Floor[Sqrt[n]]]
%t A237720 a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[(n+1)/2]}]
%t A237720 Table[a[n],{n,1,70}]
%Y A237720 Cf. A000040, A000290, A237706, A237710, A237721.
%K A237720 nonn
%O A237720 1,7
%A A237720 _Zhi-Wei Sun_, Feb 12 2014