

A219023


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


5



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


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..20000
ZhiWei Sun, Conjectures involving primes and quadratic forms, arXiv preprint arXiv:1211.1588, 2012.
Wikipedia, Oppermann's Conjecture


EXAMPLE

a(12)=2 since the 5 and 7 are the only primes p<12 with 12^212+p and 12^2+12p both prime.


MATHEMATICA

a[n_]:=a[n]=Sum[If[PrimeQ[n^2n+Prime[k]]==True&&PrimeQ[n^2+nPrime[k]]==True, 1, 0], {k, 1, PrimePi[n1]}]
Do[Print[n, " ", a[n]], {n, 1, 20000}]
Table[Total[Table[If[AllTrue[{k^2k+p, k^2+kp}, 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^2n, np=n^2+n); forprime(p=1, n1, isprime(npp) && isprime(nm+p) && c++); c} \\  M. F. Hasler, Nov 11 2012


CROSSREFS

Cf. A000040.
Sequence in context: A281116 A089233 A066620 * A025427 A245963 A291375
Adjacent sequences: A219020 A219021 A219022 * A219024 A219025 A219026


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Nov 10 2012


STATUS

approved



