A238878 a(n) = |{0 < k <= n: prime(prime(k)) - prime(k) + 1 and prime(prime(k*n)) - prime(k*n) + 1 are both prime}|.

1, 2, 3, 1, 1, 4, 3, 2, 5, 5, 3, 4, 2, 2, 3, 3, 5, 3, 1, 3, 4, 4, 2, 5, 2, 2, 7, 3, 2, 4, 4, 7, 4, 4, 4, 4, 4, 3, 4, 4, 4, 2, 4, 3, 7, 4, 9, 6, 3, 4, 5, 4, 2, 4, 4, 4, 3, 4, 5, 6, 10, 4, 4, 8, 9, 6, 5, 6, 5, 7, 8, 9, 5, 2, 5, 7, 1, 7, 4, 5

Offset: 1

Author

Zhi-Wei Sun, Mar 06 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 5, 19, 77.

(ii) For any integer n > 0, there is a number k among 1, ..., n such that 2*k + 1 and prime(prime(k^2*n)) - prime(k^2*n) + 1 are both prime.

Examples

			a(5) = 1 since prime(prime(4)) - prime(4) + 1 = prime(7) - 7 + 1 = 17 - 6 = 11 and prime(prime(4*5)) - prime(4*5) + 1 = prime(71) - 71 + 1 = 353 - 70 = 283 are both prime.
a(77) = 1 since prime(prime(3)) - prime(3) + 1 = prime(5) - 5 + 1 = 11 - 4 = 7 and prime(prime(3*77)) - prime(3*77) + 1 = prime(1453) - 1453 + 1 = 12143 - 1452 = 10691 are both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..5000
• Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Crossrefs

Cf. A000040, A234695, A237715, A238573, A238576, A238756, A238766, A238776, A238814, A238881.

Programs

Mathematica

PQ[n_]:=PrimeQ[Prime[n]-n+1]
p[k_,n_]:=PQ[Prime[k]]&&PQ[Prime[k*n]]
a[n_]:=Sum[If[p[k,n],1,0],{k,1,n}]
Table[a[n],{n,1,80}]