A232194 - OEIS

%I #8 Nov 20 2013 19:37:36

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

%T 9,3,5,4,6,3,8,4,7,5,8,3,7,4,6,3,8,3,8,2,12,4,9,4,9,4,10,3,9,7,10,5,9,

%U 4,10,4,6,5,8,3,7,5,11,7,9,8,11,5,11,8,13,4,9,5,8,7,12,6,9,5,15,7,10,5,15,10

%N Number of ways to write n = x + y (x, y > 0) with n*x + y and n*y - x both prime.

%C Conjecture: (i) a(n) > 0 for all n > 2. Also, a(n) = 1 only for n = 3, 4, 6, 20, 24.

%C (ii) Any positive integer n not among 1, 30, 54 can be written as x + y (x, y > 0) with n*x + y and n*y + x both prime.

%C (iii) Each integer n > 1 not equal to 8 can be expressed as x + y (x, y > 0) with n*x^2 + y (or x^4 + n*y) prime.

%C (iv) Any integer n > 5 can be written as p + q (q > 0) with p and n*q^2 + 1 both prime.

%C See also A232174 for a similar conjecture.

%H Zhi-Wei Sun, <a href="/A232194/b232194.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1211.1588">Conjectures involving primes and quadratic forms</a>, preprint, arXiv:1211.1588.

%e a(3) = 1 since 3 = 1 + 2 with 3*1 + 2 = 3*2 - 1 = 5 prime.

%e a(4) = 1 since 4 = 1 + 3 with 4*1 + 3 = 7 and 4*3 - 1 = 11 both prime.

%e a(6) = 1 since 6 = 1 + 5 with 6*1 + 5 = 11 and 6*5 - 1 = 29 both prime.

%e a(20) = 1 since 20 = 9 + 11 with 20*9 + 11 = 191 and 20*11 - 9 = 211 both prime.

%e a(24) = 1 since 24*19 + 5 = 461 and 24*5 - 19 = 101 both prime.

%t a[n_]:=Sum[If[PrimeQ[n*x+(n-x)]&&PrimeQ[n*(n-x)-x],1,0],{x,1,n-1}]

%t Table[a[n],{n,1,100}]

%Y Cf. A000040, A218585, A218654, A219864, A220413, A227898, A227899, A232174, A232186.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Nov 20 2013