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A231883
Number of ways to write n = x + y (x, y > 0) with x^2 + (n-2)*y^2 prime.
1
0, 0, 2, 2, 2, 2, 4, 1, 5, 2, 5, 1, 4, 4, 3, 1, 7, 2, 3, 3, 6, 7, 3, 2, 6, 2, 9, 3, 8, 3, 10, 3, 5, 8, 8, 4, 7, 5, 13, 4, 12, 6, 7, 6, 8, 10, 14, 4, 17, 9, 9, 6, 9, 5, 8, 5, 9, 7, 12, 10, 11, 7, 11, 8, 12, 4, 13, 3, 22, 6, 16, 7, 14, 8, 10, 4, 14, 4, 17, 9, 16, 6, 12, 11, 14, 4, 21, 4, 21, 8, 18, 3, 11, 14, 23, 7, 22, 5, 23, 8
OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 8, 12, 16. Moreover, if m and n are positive integers with m >= max{2, n-1} and gcd(m, n+1) = 1, then x^2 + n*y^2 is prime for some positive integers x and y with x + y = m, except for the case m = n + 3 = 29.
(ii) Let m and n be integers greater than one with m >= (n-1)/2 and gcd(m, n-1) = 1. Then x + n*y is prime for some positive integers x and y with x + y = m.
(iii) Any integer n > 3 not equal to 12 or 16 can be written as x + y (x, y > 0) with (n-2)*x - y and (n-2)*x^2 + y^2 both prime.
LINKS
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588.
EXAMPLE
a(8) = 1 since 8 = 5 + 3 with 5^2 + (8-2)*3^2 = 79 prime.
a(12) = 1 since 12 = 11 + 1 with 11^2 + (12-2)*1^2 = 131 prime.
a(16) = 1 since 16 = 15 + 1 with 15^2 + (16-2)*1^2 = 239 prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[x^2+(n-2)*(n-x)^2], 1, 0], {x, 1, n-1}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 21 2013
STATUS
approved