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A238732 Number of primes p < n with floor((n-p)/3) a square.

%I A238732 #15 Mar 04 2014 09:13:23
%S A238732 0,0,1,2,2,3,3,3,2,2,1,2,1,3,4,4,3,3,3,3,3,2,2,3,3,2,2,1,2,4,5,5,4,4,
%T A238732 3,3,1,2,2,3,3,4,3,4,4,4,2,3,2,4,5,4,3,3,5,4,4,3,3,4,4,3,3,3,4,4,3,3,
%U A238732 3,3,4,5,4,4,4,3,3,4,5,6
%N A238732 Number of primes p < n with floor((n-p)/3) a square.
%C A238732 Conjecture: (i) For any integers m > 2 and n > 2, there is a prime p < n such that floor((n-p)/m) is a square.
%C A238732 (ii) For any integer k > 2, if m and n are sufficiently large, then there is a prime p < n such that floor((n-p)/m) is a k-th power. In particular, for any integer n > 2, there is a prime p < n such that floor((n-p)/5) is a cube.
%C A238732 We have verified part (i) for m = 3, ..., 10 and n = 3, ..., 10^6.
%H A238732 Zhi-Wei Sun, <a href="/A238732/b238732.txt">Table of n, a(n) for n = 1..10000</a>
%H A238732 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.4461">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.
%e A238732 a(3) = 1 since 2 is prime with floor((3-2)/3) = 0^2.
%e A238732 a(11) = 1 since 7 is prime with floor((11-7)/3) = 1^2.
%e A238732 a(13) = 1 since 13 is prime with floor((13-11)/3) = 0^2.
%e A238732 a(28) = 1 since 23 is prime with floor((28-23)/3) = 1^2.
%e A238732 a(37) = 1 since 23 is prime with floor((37-23)/3) = 2^2.
%e A238732 a(173) = 1 since 97 is prime with floor((173-97)/3) = 5^2.
%e A238732 It seems that a(n) = 1 only for n = 3, 11, 13, 28, 37, 173.
%t A238732 SQ[n_]:=IntegerQ[Sqrt[n]]
%t A238732 s[n_,k_]:=SQ[Floor[(n-Prime[k])/3]]
%t A238732 a[n_]:=Sum[If[s[n,k],1,0],{k,1,PrimePi[n-1]}]
%t A238732 Table[a[n],{n,1,80}]
%Y A238732 Cf. A000040, A000290, A237706, A238701, A238134.
%K A238732 nonn
%O A238732 1,4
%A A238732 _Zhi-Wei Sun_, Mar 03 2014