A238573 a(n) = |{0 < k <= n: prime(k*n) + 2 is prime}|.

0, 1, 1, 0, 3, 0, 4, 1, 2, 3, 1, 3, 4, 4, 4, 4, 1, 3, 6, 5, 3, 3, 4, 6, 3, 8, 5, 6, 3, 4, 2, 10, 6, 5, 7, 8, 6, 8, 7, 5, 7, 5, 11, 7, 7, 8, 8, 11, 6, 5, 7, 11, 11, 7, 4, 9, 7, 3, 5, 7, 7, 11, 8, 13, 9, 8, 7, 7, 12, 10, 8, 11, 8, 15, 8, 9, 9, 15, 13, 4

Offset

1,5

Comments

Conjecture: (i) a(n) > 0 for all n > 6, and a(n) = 1 only for n = 2, 3, 8, 11, 17. Moreover, for any n > 0 there exists a positive integer k < 3*sqrt(n) + 6 such that prime(k*n) + 2 is prime.

(ii) For any integer n > 6, prime(k^2*n) + 2 is prime for some k = 1, ..., n.

(iii) If n > 5, then prime(k^2*(n-k)) + 2 is prime for some 0 < k < n.

Clearly, each of the three parts implies the twin prime conjecture.

We have verified part (i) of the conjecture for n up to 2*10^6.

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.

Example

a(2) = 1 since prime(1*2) + 2 = 3 + 2 = 5 is prime.
a(3) = 1 since prime(1*3) + 2 = 5 + 2 = 7 is prime.
a(8) = 1 since prime(8*8) + 2 = 311 + 2 = 313 is prime.
a(11) = 1 since prime(3*11) + 2 = 137 + 2 = 139 is prime.
a(17) = 1 since prime(1*17) + 2 = 59 + 2 = 61 is prime.

Mathematica

p[k_, n_]:=PrimeQ[Prime[k*n]+2]
a[n_]:=Sum[If[p[k, n], 1, 0], {k, 1, n}]
Table[a[n], {n, 1, 80}]

Crossrefs

Cf. A000040, A001359, A006512, A218829, A236531, A237578, A238576.

Sequence in context: A035544 A129718 A127375 * A346965 A325491 A306803

Adjacent sequences: A238570 A238571 A238572 * A238574 A238575 A238576

Keyword

nonn

Author

Zhi-Wei Sun, Mar 01 2014

Status

approved