

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
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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

ZhiWei 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, n1}]
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

ZhiWei Sun, Feb 09 2014


STATUS

approved



