

A237453


Number of primes p < n with p*n + pi(p) prime, where pi(.) is given by A000720.


6



0, 0, 1, 0, 2, 1, 1, 2, 3, 1, 1, 1, 1, 2, 3, 2, 2, 2, 2, 3, 2, 1, 2, 1, 2, 3, 3, 2, 3, 1, 1, 1, 3, 2, 4, 3, 3, 3, 2, 1, 2, 1, 1, 3, 3, 1, 2, 3, 3, 3, 4, 3, 3, 2, 2, 6, 4, 3, 5, 3, 2, 3, 2, 4, 4, 3, 1, 3, 5, 2, 5, 3, 1, 2, 3, 2, 4, 2, 3, 2
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OFFSET

1,5


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 4, and a(n) = 1 for no n > 144. Moreover, for any positive integer n, there is a prime p < sqrt(2*n)*log(5n) with p*n + pi(p) prime.
(ii) For each integer n > 8, there is a prime p <= n + 1 with (p1)*n  pi(p1) prime.
(iii) For every n = 1, 2, 3, ... there is a positive integer k < 3*sqrt(n) with k*n + prime(k) prime.
(iv) For each n > 13, there is a positive integer k < n with k*n + prime(nk) prime.
We have verified that a(n) > 0 for all n = 5, ..., 10^8.


LINKS



EXAMPLE

a(3) = 1 since 2 and 2*3 + pi(2) = 6 + 1 = 7 are both prime.
a(10) = 1 since 5 and 5*10 + pi(5) = 50 + 3 = 53 are both prime.
a(107) = 1 since 89 and 89*107 + pi(89) = 9523 + 24 = 9547 are both prime.
a(144) = 1 since 59 and 59*144 + pi(59) = 8496 + 17 = 8513 are both prime.


MATHEMATICA

a[n_]:=Sum[If[PrimeQ[Prime[k]*n+k], 1, 0], {k, 1, PrimePi[n1]}]
Table[a[n], {n, 1, 80}]


PROG

(PARI) vector(100, n, sum(k=1, primepi(n1), isprime(prime(k)*n+k))) \\ Colin Barker, Feb 08 2014


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



