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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. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,24

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.

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.

EXAMPLE

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.

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}]

CROSSREFS

Cf. A002375, A046927, A219055, A219157, A218867, A219185, A218754, A218825, A219052.

Sequence in context: A096271 A285640 A231189 * A279210 A232243 A034876

Adjacent sequences:  A219555 A219556 A219557 * A219559 A219560 A219561

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Nov 23 2012

STATUS

approved

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Last modified February 19 09:13 EST 2018. Contains 299330 sequences. (Running on oeis4.)