login
A238568
a(n) = |{0 < k < n: n^2 - pi(k*n) is prime}|, where pi(x) denotes the number of primes not exceeding x.
2
0, 1, 1, 1, 2, 2, 2, 1, 2, 1, 3, 2, 4, 3, 4, 2, 2, 5, 5, 3, 4, 4, 8, 1, 3, 3, 4, 3, 4, 3, 6, 3, 4, 4, 3, 4, 6, 3, 5, 2, 1, 8, 3, 10, 6, 5, 5, 9, 7, 6, 3, 8, 7, 9, 2, 5, 5, 2, 2, 9, 7, 3, 5, 8, 7, 6, 8, 7, 9, 9, 6, 3, 7, 8, 14, 5, 9, 10, 8, 11
OFFSET
1,5
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 8, 10, 24, 41.
(ii) For any integer n > 6, there is a positive integer k < n with n^2 + pi(k*n) - 1 prime.
(iii) If n > 2, then pi(n^2) - pi(k*n) is prime for some 0 < k < n. If n > 1, then pi(n^2) + pi(k*n) - 1 is prime for some 0 < k < n.
LINKS
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014-2016.
EXAMPLE
a(2) = 1 since 2^2 - pi(1*2) = 4 - 1 = 3 is prime.
a(3) = 1 since 3^2 - pi(1*3) = 9 - 2 = 7 is prime.
a(4) = 1 since 4^2 - pi(3*4) = 16 - 5 = 11 is prime.
a(8) = 1 since 8^2 - pi(4*8) = 64 - 11 = 53 is prime.
a(10) = 1 since 10^2 - pi(6*10) = 100 - 17 = 83 is prime.
a(24) = 1 since 24^2 - pi(14*24) = 576 - 67 = 509 is prime.
a(41) = 1 since 41^2 - pi(10*41) = 1681 - 80 = 1601 is prime.
MATHEMATICA
p[k_, n_]:=PrimeQ[n^2-PrimePi[k*n]]
a[n_]:=Sum[If[p[k, n], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 28 2014
STATUS
approved