A234694 a(n) = |{0 < k < n: p = k + prime(n-k) and prime(p) - p + 1 are both prime}|.

0, 1, 0, 2, 1, 2, 1, 0, 0, 2, 2, 4, 1, 1, 2, 4, 2, 1, 1, 2, 3, 3, 2, 3, 1, 1, 1, 3, 5, 4, 3, 4, 3, 3, 3, 2, 4, 3, 2, 5, 4, 4, 4, 1, 1, 5, 4, 2, 1, 2, 5, 5, 2, 3, 4, 2, 3, 5, 7, 7, 6, 2, 5, 6, 2, 5, 4, 4, 7, 6, 6, 5, 4, 8, 7, 4, 5, 3, 5, 7, 3, 5, 4, 7, 6, 7, 2

Offset: 1

Author

Zhi-Wei Sun, Dec 29 2013

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 9. Also, for any integer n > 51 there is a positive integer k < n such that p = k + prime(n-k) and prime(p) + p + 1 are both prime.

(ii) If n > 9 (or n > 21), then there is a positive integer k < n such that m - 1 and prime(m) + m (or prime(m) - m, resp.) are both prime, where m = k + prime(n-k).

(iii) If n > 483, then for some 0 < k < n both prime(m) + m and prime(m) - m are prime, where m = k + prime(n-k).

(iv) If n > 3, then there is a positive integer k < n such that prime(k + prime(n-k)) + 2 is prime.

Clearly, part (i) of the conjecture implies that there are infinitely many primes p with prime(p) - p + 1 (or prime(p) + p + 1) also prime.

See A234695 for primes p with prime(p) - p + 1 also prime.

Examples

			a(5) = 1 since 2 + prime(3) = 7 and prime(7) - 6 = 11 are both prime.
a(25) = 1 since 20 + prime(5) = 31 and prime(31) - 30 = 97 are both prime.
a(27) = 1 since 18 + prime(9) = 41 and prime(41) - 40 = 139 are both prime.
a(45) = 1 since 6 + prime(39) = 173 and prime(173) - 172 = 859 are both prime.
a(49) = 1 since 26 + prime(23) = 109 and prime(109) - 108 = 491 are both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Crossrefs

Cf. A000040, A014688, A014689, A014692, A064269, A064270, A232861, A233150, A233183, A233206, A233296, A234695.

Programs

Mathematica

f[n_,k_]:=k+Prime[n-k]
q[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[Prime[f[n,k]]-f[n,k]+1]
a[n_]:=Sum[If[q[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]