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A219558 Number of odd prime pairs {p,q} (p>q) such that p+(1+(n mod 2))q=n and ((p-1-(n mod 2))/q)=((q+1)/p)=1 where (-) denotes the Legendre symbol.

%I A219558 #11 Jan 27 2014 09:16:05
%S A219558 0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,2,0,1,0,0,0,1,1,1,1,0,
%T A219558 0,0,0,1,0,0,0,0,0,2,1,0,0,3,0,2,0,1,1,1,1,2,2,0,0,0,0,1,1,0,0,1,0,0,
%U A219558 1,1,0,3,0,1,1,1,0,1,1,2,1,1,2,0,0,2,1,2,1,1,0,1,1,2,2,3,0,0,0,0
%N A219558 Number of odd prime pairs {p,q} (p>q) such that p+(1+(n mod 2))q=n and ((p-1-(n mod 2))/q)=((q+1)/p)=1 where (-) denotes the Legendre symbol.
%C A219558 For any integer m, define s(m) as the smallest positive integer s such that for each n=s,s+1,... there are primes p>q>2 with p+(1+(n mod 2))q=n and ((p-(1+(n mod 2))m)/q)=((q+m)/p)=1. If such a positive integer s does not exist, then we set s(m)=0.
%C A219558 Zhi-Wei Sun has the following general conjecture: s(m) is always positive. In particular, s(0)=1239,
%C A219558   s(1)=1470, s(-1)=2192, s(2)=1034, s(-2)=1292,
%C A219558   s(3)=1698, s(-3)=1788, s(4)=848, s(-4)=1458,
%C A219558   s(5)=1490, s(-5)=2558, s(6)=1115, s(-6)=1572,
%C A219558   s(7)=1550, s(-7)=932,  s(8)=825, s(-8)=2132,
%C A219558   s(9)=1154, s(-9)=1968, s(10)=1880, s(-10)=1305,
%C A219558   s(11)=1052, s(-11)=1230, s(12)=2340, s(-12)=1428,
%C A219558   s(13)=2492, s(-13)=2673, s(14)=1412, s(-14)=1638,
%C A219558   s(15)=1185, s(-15)=1230, s(16)=978, s(-16)=1605,
%C A219558   s(17)=1154, s(-17)=1692, s(18)=1757, s(-18)=2292,
%C A219558   s(19)=1230, s(-19)=2187, s(20)=2048, s(-20)=1372,
%C A219558   s(21)=1934, s(-21)=1890, s(22)=1440, s(-22)=1034,
%C A219558   s(23)=1964, s(-23)=1322, s(24)=1428, s(-24)=2042,
%C A219558   s(25)=1734, s(-25)=1214, s(26)=1260, s(-26)=1230,
%C A219558   s(27)=1680, s(-27)=1154, s(28)=1652, s(-28)=1808,
%C A219558   s(29)=1112, s(-29)=1670, s(30)=1820, s(-30)=1284.
%C A219558 Note that s(1)=1470 means that a(n)>0 for all n=1470,1471,... That s(0)=1239 is related to a conjecture of _Olivier Gérard_ and Zhi-Wei Sun.
%C A219558 If we replace ((p-1-(n mod 2))/q)=((q+1)/p)=1 in the definition of a(n) by ((p-1)/q)=((q+1)/p)=1, then the new a(n) seems positive for any n>1181.
%H A219558 Zhi-Wei Sun, <a href="/A219558/b219558.txt">Table of n, a(n) for n = 1..10000</a>
%H A219558 Olivier Gérard and Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;c08d598.1211">Refining Goldbach's conjecture by using quadratic residues</a>, a message to Number Theory List, Nov. 19, 2012.
%H A219558 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1211.1588">Conjectures involving primes and quadratic forms</a>, arXiv:1211.1588.
%e A219558 a(14)=1 since 14=11+3 with ((11-1)/3)=((3+1)/11)=1.
%e A219558 a(31)=1 since 31=17+2*7 with ((17-2)/7)=((7+1)/17)=1.
%t A219558 a[n_]:=a[n]=Sum[If[PrimeQ[n-(1+Mod[n,2])Prime[k]]==True&&JacobiSymbol[n-(1+Mod[n,2])(Prime[k]+1),Prime[k]]==1&&JacobiSymbol[Prime[k]+1,n-(1+Mod[n,2])Prime[k]]==1,1,0],{k,2,PrimePi[(n-1)/(2+Mod[n,2])]}]
%t A219558 Do[Print[n," ",a[n]],{n,1,10000}]
%Y A219558 Cf. A002375, A046927, A219055, A219157, A218867, A219185, A218754, A218825, A219052.
%K A219558 nonn
%O A219558 1,24
%A A219558 _Zhi-Wei Sun_, Nov 23 2012