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%I #12 May 10 2019 04:35:33
%S 0,0,0,1,0,0,2,1,1,3,3,3,1,2,2,5,2,4,6,4,3,3,6,7,4,4,5,2,5,7,5,8,3,7,
%T 7,6,6,10,6,12,8,7,8,12,7,9,14,9,6,8,10,7,10,13,9,12,11,12,16,12,12,
%U 13,10,13,14,13,12,14,13,13,16,12,13,20,16,11,12,13,12,19,18,12,17,21,12,19,17,11,19,17,18,18,20,12,23,17,13,18,18,14
%N Number of ways to write 2*n+1 as p + q + r with 2*p + 4*q + 6*r a square, where p,q,r are odd primes.
%C 2-4-6 Conjecture: a(n) > 0 for all n > 6. In other words, any odd integer greater than 14 can be written as the sum of three odd primes p,q,r for which 2*p + 4*q + 6*r is an integer square.
%C This is stronger than the solved weak Goldbach conjecture (A068307), and it is motivated by the author's 1-3-5 conjecture in A271518.
%C We have verified a(n) > 0 for all n = 7..3*10^5.
%H Zhi-Wei Sun, <a href="/A308028/b308028.txt">Table of n, a(n) for n = 1..2000</a>
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190.
%H Zhi-Wei Sun, <a href="https://mathoverflow.net/questions/331170">Write 2n+1 > 14 as p+q+r with p,q,r odd primes and 2p+4q+6r a square</a>, Question 331170 on MathOverflow, May 10, 2019.
%e a(8) = 1 with 2*8+1 = 17 = 7 + 5 + 5 and 2*7 + 4*5 + 6*5 = 8^2.
%e a(9) = 1 with 2*9+1 = 19 = 11 + 3 + 5 and 2*11 + 4*3 + 6*5 = 8^2.
%e a(13) = 1 with 2*13+1 = 7 + 17 + 3 and 2*7 + 4*17 + 6*3 = 10^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];p[n_]:=p[n]=Prime[n];
%t tab={};Do[r=0;Do[If[PrimeQ[2n+1-p[i]-p[j]]&&SQ[2p[i]+4p[j]+6(2n+1-p[i]-p[j])],r=r+1],{i,2,PrimePi[2n]},{j,2,PrimePi[2n-p[i]]}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000040, A068307, A271518.
%K nonn
%O 1,7
%A _Zhi-Wei Sun_, May 09 2019
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