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A237582 a(n) = |{0 < k < n: pi(n + k^2) is prime}|, where pi(.) is given by A000720. 3
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 (list; graph; refs; listen; history; text; internal format)
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: A305866 A257396 A293519 * A097352 A076050 A130799

Adjacent sequences:  A237579 A237580 A237581 * A237583 A237584 A237585

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Feb 09 2014

STATUS

approved

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Last modified October 21 14:57 EDT 2018. Contains 316424 sequences. (Running on oeis4.)