A237615 - OEIS

A237615

a(n) = |{0 < k < n: k^2 + k - 1 and pi(k*n) are both prime}|, where pi(.) is given by A000720.

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,6

COMMENTS

Conjecture: (i) a(n) > 0 for all n > 5.

(ii) For each n = 4, 5, ..., there is a positive integer k < n with k^2 + k - 1 and pi(k*n) + 1 both prime. Also, for any integer n > 6, there is a positive integer k < n with k^2 + k - 1 and pi(k*n) - 1 both prime.

(iii) For every integer n > 15, there is a positive integer k < n such that pi(k) - 1 and pi(k*n) are both prime.

Note that part (i) is a refinement of the first assertion in the comments in A237578.

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..5000

Zhi-Wei Sun, A combinatorial conjecture on primes, a message to Number Theory List, Feb. 9, 2014.

EXAMPLE

a(8) = 1 since 4^2 + 4 - 1 = 19 and pi(4*8) = 11 are both prime.

a(33) = 1 since 28^2 + 28 - 1 = 811 and pi(28*33) = 157 are both prime.

MATHEMATICA

p[k_, n_]:=PrimeQ[k^2+k-1]&&PrimeQ[PrimePi[k*n]]

a[n_]:=Sum[If[p[k, n], 1, 0], {k, 1, n-1}]

Table[a[n], {n, 1, 70}]