A237720 Number of primes p <= (n+1)/2 with floor( sqrt(n-p) ) prime.

0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 3, 2, 2, 2, 2, 1, 1, 2, 2, 2, 3, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 4, 4, 3, 4, 3, 4, 4, 4, 3, 4, 3, 3, 4, 5, 4, 5, 4, 5, 6, 6, 5, 6, 7, 8, 8, 8, 7, 7, 5, 6, 5, 5

Offset

1,7

Comments

Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 6, 23, 24, 111, 112, ..., 120.

(ii) For any integer n > 2, there is a prime p < n with floor(sqrt(n+p)) prime.

Note that floor(sqrt(n)) is the number of squares among 1, ..., n.

See also A237705, A237706 and A237721 for similar conjectures.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Example

a(6) = 1 since 2 and floor(sqrt(6-2)) = 2 are both prime.
a(23) = 1 since 11 and floor(sqrt(23-11)) = 3 are both prime.
a(24) = 1 since 11 and floor(sqrt(24-11)) = 3 are both prime.
a(27) = 2 since 2 and floor(sqrt(27-2)) = 5 are both prime, and 13 and floor(sqrt(27-13)) = 3 are both prime.
a(n) = 1 for n = 111, ..., 116 since 53 and floor(sqrt(n-53)) = 7 are both prime.
a(n) = 1 for n = 117, 118, 119, 120 since 59 and floor(sqrt(n-59)) = 7 are both prime.

Mathematica

q[n_]:=PrimeQ[Floor[Sqrt[n]]]
a[n_]:=Sum[If[q[n-Prime[k]], 1, 0], {k, 1, PrimePi[(n+1)/2]}]
Table[a[n], {n, 1, 70}]

Crossrefs

Cf. A000040, A000290, A237706, A237710, A237721.

Sequence in context: A071673 A174199 A072660 * A075172 A347643 A237261

Adjacent sequences: A237717 A237718 A237719 * A237721 A237722 A237723

Keyword

nonn

Author

Zhi-Wei Sun, Feb 12 2014

Status

approved