OFFSET
1,3
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 1. Also, a(n) = 1 only for n = 2, 5, 8, 14, 19, 20, 24, 32, 54, 68, 101, 168.
(ii) Every n = 3, 4, ... can be written as x + y (x, y > 0) with x*n + y and x*n - y both prime.
(iii) Any integer n > 2 can be written as P + q (q > 0) with p and p + n*q both prime. Also, any integer n > 7 can be written as p + q (q > 0) with p and n*q - p both prime.
In a paper published in 2017, the author announced a USD $200 prize for the first solution to his conjecture that a(n) > 0 for all n > 1. - Zhi-Wei Sun, Dec 03 2017
REFERENCES
D. A. Cox, Primes of the Form x^2 + n*y^2, John Wiley & Sons, 1989.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II: CANT, New York, NY, USA, 2015 and 2016, Springer Proc. in Math. & Stat., Vol. 220, Springer, New York, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT].)
EXAMPLE
a(2) = 1 since 2 = 1 + 1 with 1 + 2*1 = 1^2 + 2*1^2 = 3 prime.
a(5) = 1 since 5 = 3 + 2 with 3 + 5*2 = 13 and 3^2 + 5*2^2 = 29 both prime.
a(8) = 1 since 8 = 5 + 3 with 5 + 8*3 = 29 and 5^2 + 8*3^2 = 97 both prime.
a(14) = 1 since 14 = 9 + 5 with 9 + 14*5 = 79 and 9^2 + 14*5^2 = 431 both prime.
a(19) = 1 since 19 = 13 + 6 with 13 + 19*6 = 127 and 13^2 + 19*6^2 = 853 both prime.
a(20) = 1 since 20 = 11 + 9 with 11 + 20*9 = 191 and 11^2 + 20*9^2 = 1741 both prime.
a(24) = 1 since 24 = 5 + 19 with 5 + 24*19 = 461 and 5^2 + 24*19^2 = 8689 both prime.
a(32) = 1 since 32 = 23 + 9 with 23 + 32*9 = 311 and 23^2 + 32*9^2 = 3121 both prime.
a(54) = 1 since 54 = 35 + 19 with 35 + 54*19 = 1061 and 35^2 + 54*19^2 = 20719 both prime.
a(68) = 1 since 68 = 45 + 23 with 45 + 68*23 = 1609 and 45^2 + 68*23^2 = 37997 both prime.
a(101) = 1 since 101 = 98 + 3 with 98 + 101*3 = 401 and 98^2 + 101*3^2 = 10513 both prime.
a(168) = 1 since 168 = 125 + 43 with 125 + 168*43 = 7349 and 125^2 + 168*43^2 = 326257 both prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[k+n(n-k)]&&PrimeQ[k^2+n(n-k)^2], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 19 2013
STATUS
approved