|
%I #9 Nov 22 2013 01:38:40
%S 1,3,1,2,3,2,1,6,4,1,4,6,3,8,1,1,6,1,1,9,2,4,5,3,1,2,7,4,5,8,1,12,4,4,
%T 12,3,4,9,10,1,5,9,5,11,7,4,9,2,4,19,1,1,14,4,6,16,8,5,8,7,2,11,8,1,
%U 16,3,5,9,4,3,8,8,6,16,4,3,12,13,5,11,5,3,10,10,7,12,7,4,17,20,1,17,5,6,15,4,5,18,5,7
%N Number of ways to write 2*n + 1 = x + y (x, y > 0) with x^3 + y^2 and x^2 + y^2 both prime.
%C Conjecture: a(n) > 0 for all n > 0. Also, any odd integer greater than one can be written as x + y (0 < x < y) with x^3 + y^2 prime.
%C The conjecture implies that there are infinitely many primes of the form x^3 + y^2 (x, y > 0) with x^2 + y^2 also prime.
%C Note that Ming-Zhi Zhang ever asked (cf. A036468) whether any odd integer greater than one can be written as x + y (x, y > 0) with x^2 + y^2 prime.
%H Zhi-Wei Sun, <a href="/A232269/b232269.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(10) = 1 since 2*10 + 1 = 1 + 20 with 1^2 + 20^2 = 1^3 + 20^2 = 401 prime.
%e a(15) = 1 since 2*15 + 1 = 25 + 6 with 25^2 + 6^2 = 661 and 25^3 + 6^2 = 15661 both prime.
%e a(40) = 1 since 2*40 + 1 = 55 + 26 with 55^2 + 26^2 = 3701 and 55^3 + 26^2 = 167051 both prime.
%e a(91) =1 since 2*91 + 1 = 85 + 98 with 85^2 + 98^2 = 16829 and 85^3 + 98^2 = 623729 both prime.
%t a[n_]:=Sum[If[PrimeQ[x^3+(2n+1-x)^2]&&PrimeQ[x^2+(2n+1-x)^2],1,0],{x,1,2n}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000040, A036468, A066649, A220413, A232174, A232186.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Nov 22 2013
|
