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