cp's OEIS Frontend

A237721 Number of primes p <= n with floor( sqrt(n-p) ) a square.

%I A237721 #12 Feb 12 2014 06:43:30
%S A237721 0,1,2,2,3,2,2,2,1,1,1,1,2,2,1,1,1,2,4,4,4,4,5,5,5,5,4,3,5,4,5,4,4,4,
%T A237721 4,3,4,3,4,4,4,3,4,3,3,3,4,3,3,3,2,2,4,3,3,2,2,2,4,4,5,4,4,4,3,2,3,2,
%U A237721 3,3
%N A237721 Number of primes p <= n with floor( sqrt(n-p) ) a square.
%C A237721 Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 9, 10, 11, 12, 15, 16, 17.
%C A237721 We have verified this for n up to 10^6.
%C A237721 See also A237705, A237706 and A237720 for similar conjectures.
%H A237721 Zhi-Wei Sun, <a href="/A237721/b237721.txt">Table of n, a(n) for n = 1..10000</a>
%e A237721 a(2) = 1 since 2 is prime with floor(sqrt(2-2)) = 0^2.
%e A237721 a(3) = 2 since 2 is prime with floor(sqrt(3-2)) = 1^2, and 3 is prime with floor(sqrt(3-3)) = 0^2.
%e A237721 a(9) = a(10) = 1 since 7 is prime with floor(sqrt(9-7)) = floor(sqrt(10-7)) = 1^2.
%e A237721 a(11) = 1 since 11 is prime with floor(sqrt(11-11)) = 0^2.
%e A237721 a(12) = 1 since 11 is prime with floor(sqrt(12-11)) = 1^2.
%e A237721 a(15) = a(16) = 1 since 13 is prime with floor(sqrt(15-13)) = floor(sqrt(16-13)) = 1^2.
%e A237721 a(17) = 1 since 17 is prime with floor(sqrt(17-17)) = 0^2.
%t A237721 SQ[n_]:=IntegerQ[Sqrt[n]]
%t A237721 q[n_]:=SQ[Floor[Sqrt[n]]]
%t A237721 a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[n]}]
%t A237721 Table[a[n],{n,1,70}]
%Y A237721 Cf. A000040, A000290, A237705, A237706, A237710, A237720.
%K A237721 nonn
%O A237721 1,3
%A A237721 _Zhi-Wei Sun_, Feb 12 2014