

A218825


Number of ways to write 2n1 as p+2q with p, q and p^2+60q^2 all prime.


10



0, 0, 0, 0, 0, 0, 0, 0, 3, 1, 1, 3, 1, 3, 3, 1, 2, 2, 1, 2, 3, 1, 2, 3, 1, 2, 2, 1, 3, 1, 1, 3, 3, 4, 3, 1, 2, 5, 3, 1, 3, 2, 4, 3, 3, 1, 7, 4, 1, 5, 3, 5, 8, 4, 3, 4, 3, 3, 5, 4, 4, 3, 2, 3, 5, 3, 5, 7, 3, 2, 9, 4, 4, 6, 3, 3, 8, 6, 1, 4, 5, 2, 7, 1, 4, 2, 4, 5, 5, 2, 4, 4, 3, 2, 5, 4, 5, 6, 4, 1
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OFFSET

1,9


COMMENTS

Conjecture: a(n)>0 for all n>8.
This conjecture is stronger than Lemoine's conjecture. It has been verified for n up to 10^8.
ZhiWei Sun also made the following general conjecture: For any positive integer n, the set E(n) of positive odd integers not of the form p+2q with p, q, p^2+4(2^n1)q^2 all prime, is finite. In particular, if we let M(n) denote the maximal element of E(n), then M(1)=3449, M(2)=1711, E(3)={1,3,5,7,31,73}, E(4)={1,3,5,7,9,11,13,15},
M(5)=6227, M(6)=1051, M(7)=2239, M(8)=2599, M(9)=7723,
M(10)=781, M(11)=1163, M(12)=587, M(13)=11443,
M(14)=2279, M(15)=157, M(16)=587, M(17)=32041,
M(18)=1051, M(19)=2083, M(20)=4681.
Conjecture verified for 2n1 up to 10^9 for n <= 4 and up to 10^6 for n <= 20.  Mauro Fiorentini, Jul 20 2023
ZhiWei Sun also guessed that for any positive even integer d not congruent to 2 modulo 6 there exists a prime p(d) such that for any prime p>p(d) there is a prime q<p with p^2+dq^2 prime. In particular, we may take
p(4)=p(6)=3, p(10)=5, p(12)=3, p(16)=2, p(18)=3,
p(22)=11, p(24)=17, and p(28)=p(30)=7.


LINKS



EXAMPLE

a(10)=1 since the only primes p and q with p^2+60q^2 prime and p+2q=19 are p=13 and q=3.


MATHEMATICA

a[n_]:=a[n]=Sum[If[PrimeQ[q]==True&&PrimeQ[2n12q]==True&&PrimeQ[(2n12q)^2+60q^2]==True, 1, 0], {q, 1, n1}]
Do[Print[n, " ", a[n]], {n, 1, 20000}]


PROG

(PARI) A218825(n)={my(c=0, n21=n*21); forprime(q=2, n1, isprime(n212*q)  next; isprime(q^2*60+(n212*q)^2) && c++); c} \\ M. F. Hasler, Nov 07 2012


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



