A238703 a(n) = |{0 < k < n: floor(k*n/3) is prime}|.

0, 0, 1, 1, 2, 1, 3, 3, 1, 3, 4, 0, 4, 2, 1, 3, 5, 0, 4, 4, 1, 4, 5, 0, 3, 4, 0, 3, 6, 0, 5, 4, 1, 6, 6, 0, 7, 4, 1, 5, 4, 0, 7, 6, 0, 8, 5, 0, 8, 7, 1, 6, 7, 0, 9, 9, 1, 9, 8, 0, 6, 7, 0, 7, 12, 0, 9, 7, 1, 11, 10, 0, 6, 8, 0, 7, 9, 0, 7, 12

Offset: 1

Author

Zhi-Wei Sun, Mar 03 2014

Keywords

• nonn

Comments

Conjecture: If n > m > 0 with n not divisible by m, then floor(k*n/m) is prime for some 0 < k < n.

Examples

			a(4) = 1 since floor(2*4/3) = 2 is prime.
If p is a prime, then a(3*p) = 1 since floor(k*3p/3) = k*p is prime only for k = 1. If m > 1 is composite, then a(3*m) = 0 since floor(k*3m/3) = k*m is composite for all k > 0.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..6000
• Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Crossrefs

Cf. A000040, A237578, A237615, A238645, A238701.

Programs

Mathematica

p[n_,k_]:=PrimeQ[Floor[k*n/3]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,80}]