A219025
Number of primes p
0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 1, 2, 4, 1, 2, 1, 3, 2, 2, 2, 2, 3, 2, 3, 1, 1, 2, 5, 2, 2, 2, 4, 3, 3, 4, 1, 2, 5, 3, 2, 2, 5, 4, 1, 3, 1, 3, 5, 3, 3, 3, 3, 4, 4, 2, 6, 4, 7, 5, 2, 3, 3, 7, 5, 3, 5, 5, 7, 4, 4, 2, 3, 4, 2, 3, 3, 6, 6, 3, 2, 5, 4, 7, 3, 4, 2, 3, 7, 1, 6, 4, 5, 6
Offset: 1
Keywords
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.
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, 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, 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
Offset: 1
Keywords
Comments
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.
Zhi-Wei Sun has the following general conjecture: s(m) is always positive. In particular, s(0)=1239,
s(1)=1470, s(-1)=2192, s(2)=1034, s(-2)=1292,
s(3)=1698, s(-3)=1788, s(4)=848, s(-4)=1458,
s(5)=1490, s(-5)=2558, s(6)=1115, s(-6)=1572,
s(7)=1550, s(-7)=932, s(8)=825, s(-8)=2132,
s(9)=1154, s(-9)=1968, s(10)=1880, s(-10)=1305,
s(11)=1052, s(-11)=1230, s(12)=2340, s(-12)=1428,
s(13)=2492, s(-13)=2673, s(14)=1412, s(-14)=1638,
s(15)=1185, s(-15)=1230, s(16)=978, s(-16)=1605,
s(17)=1154, s(-17)=1692, s(18)=1757, s(-18)=2292,
s(19)=1230, s(-19)=2187, s(20)=2048, s(-20)=1372,
s(21)=1934, s(-21)=1890, s(22)=1440, s(-22)=1034,
s(23)=1964, s(-23)=1322, s(24)=1428, s(-24)=2042,
s(25)=1734, s(-25)=1214, s(26)=1260, s(-26)=1230,
s(27)=1680, s(-27)=1154, s(28)=1652, s(-28)=1808,
s(29)=1112, s(-29)=1670, s(30)=1820, s(-30)=1284.
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.
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.
Examples
a(14)=1 since 14=11+3 with ((11-1)/3)=((3+1)/11)=1. a(31)=1 since 31=17+2*7 with ((17-2)/7)=((7+1)/17)=1.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
- Olivier Gérard and Zhi-Wei Sun, Refining Goldbach's conjecture by using quadratic residues, a message to Number Theory List, Nov. 19, 2012.
- Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.
Programs
-
Mathematica
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])]}] Do[Print[n," ",a[n]],{n,1,10000}]
Comments
Examples
Links
Crossrefs
Programs
Mathematica
a[n_]:=a[n]=Sum[If[PrimeQ[6n-Prime[k]]==True&&PrimeQ[6n+Prime[k]]==True,1,0],{k,1,PrimePi[n-1]}] Do[Print[n," ",a[n]],{n,1,20000}]