A237582 a(n) = |{0 < k < n: pi(n + k^2) is prime}|, where pi(.) is given by A000720.

0, 1, 1, 1, 1, 1, 2, 3, 2, 3, 4, 1, 2, 2, 3, 6, 6, 5, 5, 5, 5, 6, 7, 7, 6, 5, 6, 5, 6, 7, 8, 9, 8, 10, 9, 8, 6, 6, 6, 6, 7, 9, 9, 10, 11, 11, 13, 11, 9, 9, 10, 10, 8, 6, 6, 5, 4, 8, 9, 10, 12, 11, 14, 15, 15, 15, 12, 14, 15, 17, 16, 13, 11, 11, 13, 16, 18, 24, 25, 20

Offset

1,7

Comments

Conjecture: (i) For each a = 2, 3, ... there is a positive integer N(a) such that for any integer n > N(a) there is a positive integer k < n with pi(n + k^a) prime. In particular, we may take (N(2), N(3), ..., N(9)) = (1, 1, 9, 26, 8, 9, 18, 1).

(ii) If n > 6, then pi(n^2 + k^2) is prime for some 0 < k < n. If n > 27, then pi(n^3 + k^3) is prime for some 0 < k < n. In general, for each a = 2, 3, ..., if n is sufficiently large then pi(n^a + k^a) is prime for some 0 < k < n.

For any integer n > 1, it is easy to show that pi(n + k) is prime for some 0 < k < n.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000

Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Example

a(5) = 1 since pi(5 + 1^2) = 3 is prime.
a(6) = 1 since pi(6 + 5^2) = pi(31) = 11 is prime.
a(9) = 2 since pi(9 + 3^2) = pi(18) = 7 and pi(9 + 5^2) = pi(34) = 11 are both prime.
a(12) = 1 since pi(12 + 10^2) = pi(112) = 29 is prime.

Mathematica

p[n_]:=PrimeQ[PrimePi[n]]
a[n_]:=Sum[If[p[n+k^2], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 80}]

Crossrefs

Cf. A000040, A000720, A185636, A237453, A237496, A237497, A237578.

Sequence in context: A328406 A257396 A293519 * A097352 A076050 A130799

Adjacent sequences: A237579 A237580 A237581 * A237583 A237584 A237585

Keyword

nonn

Author

Zhi-Wei Sun, Feb 09 2014

Status

approved