A219023 Number of primes p<n such that n^2-n+p and n^2+n-p are both prime.

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 2, 0, 1, 1, 0, 0, 2, 1, 0, 2, 0, 0, 0, 2, 1, 1, 0, 2, 1, 0, 2, 3, 0, 2, 2, 0, 1, 4, 1, 2, 1, 0, 0, 3, 1, 1, 3, 0, 0, 1, 2, 1, 1, 1, 1, 0, 0, 2, 3, 1, 0, 3, 1, 2, 1, 0, 1, 4, 0, 1, 2, 0, 2, 3, 0, 0, 4, 0, 2, 2, 0, 1, 3, 2, 1, 4, 1, 1, 3, 3, 2, 3, 1, 2, 1, 0, 2, 4, 2

Offset

1,12

Comments

Conjecture: a(n)>0 for all n>2732.

We have verified this conjecture for n up to 1.4*10^7. Note that the conjecture is stronger than Oppermann's conjecture which states that for any integer n>1 both of the two intervals (n^2-n,n^2) and (n^2,n^2+n) contain primes.

Zhi-Wei Sun also made the following conjectures: For n>3512 there is a prime p in (n,2n) such that both n^2-n+p and n^2+n-p are prime. For n>1828 there is a prime p<n such that both n^2-n-p and n^2+n+p are prime. For n>4517 there is a prime in (n,2n) such that both n^2-n-p and n^2+n+p are prime.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..20000

Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv preprint arXiv:1211.1588 [math.NT], 2012-2017.

Wikipedia, Oppermann's conjecture

Example

a(12)=2 since the 5 and 7 are the only primes p<12 with 12^2-12+p and 12^2+12-p both prime.

Mathematica

a[n_]:=a[n]=Sum[If[PrimeQ[n^2-n+Prime[k]]==True&&PrimeQ[n^2+n-Prime[k]]==True, 1, 0], {k, 1, PrimePi[n-1]}]
Do[Print[n, " ", a[n]], {n, 1, 20000}]
Table[Total[Table[If[AllTrue[{k^2-k+p, k^2+k-p}, PrimeQ], 1, 0], {p, Prime[ Range[ PrimePi[k]]]}]], {k, 100}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 23 2017 *)

Prog

(PARI) A219023(n)={my(c=0, nm=n^2-n, np=n^2+n); forprime(p=1, n-1, isprime(np-p) && isprime(nm+p) && c++); c} \\ - M. F. Hasler, Nov 11 2012

Crossrefs

Cf. A000040.

Sequence in context: A089233 A343653 A066620 * A025427 A348536 A245963

Adjacent sequences: A219020 A219021 A219022 * A219024 A219025 A219026

Keyword

nonn

Author

Zhi-Wei Sun, Nov 10 2012

Status

approved