A238732 - OEIS

%I #15 Mar 04 2014 09:13:23

%S 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 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 3,3,4,5,4,4,4,3,3,4,5,6

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

%C 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 (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 We have verified part (i) for m = 3, ..., 10 and n = 3, ..., 10^6.

%H Zhi-Wei Sun, <a href="/A238732/b238732.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1402.4461">Problems on combinatorial properties of primes</a>, arXiv:1402.6641, 2014.

%e a(3) = 1 since 2 is prime with floor((3-2)/3) = 0^2.

%e a(11) = 1 since 7 is prime with floor((11-7)/3) = 1^2.

%e a(13) = 1 since 13 is prime with floor((13-11)/3) = 0^2.

%e a(28) = 1 since 23 is prime with floor((28-23)/3) = 1^2.

%e a(37) = 1 since 23 is prime with floor((37-23)/3) = 2^2.

%e a(173) = 1 since 97 is prime with floor((173-97)/3) = 5^2.

%e It seems that a(n) = 1 only for n = 3, 11, 13, 28, 37, 173.

%t SQ[n_]:=IntegerQ[Sqrt[n]]

%t s[n_,k_]:=SQ[Floor[(n-Prime[k])/3]]

%t a[n_]:=Sum[If[s[n,k],1,0],{k,1,PrimePi[n-1]}]

%t Table[a[n],{n,1,80}]

%Y Cf. A000040, A000290, A237706, A238701, A238134.

%K nonn

%O 1,4

%A _Zhi-Wei Sun_, Mar 03 2014