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A192056 a(n) = |{0<k<n/2: k^2+(n-k)^2-3(n-1 mod 2) is prime and JacobiSymbol[k,n+3(n-1 mod 2)]=1}| 1
0, 0, 1, 1, 1, 2, 2, 2, 2, 1, 3, 1, 2, 2, 3, 3, 2, 1, 3, 4, 2, 6, 2, 1, 8, 3, 3, 6, 2, 1, 3, 3, 1, 5, 7, 5, 4, 4, 3, 3, 6, 3, 3, 6, 3, 5, 3, 7, 5, 7, 6, 4, 5, 1, 8, 8, 2, 4, 6, 1, 5, 2, 4, 9, 8, 3, 6, 7, 3, 5, 5, 5, 3, 3, 5, 9, 4, 13, 6, 5, 9, 7, 7, 3, 10, 9, 8, 9, 7, 4, 7, 13, 5, 7, 10, 4, 4, 11, 4, 5 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,6

COMMENTS

Conjecture: a(n)>0 for all n>2. Moreover, if n>2 is not among 12, 18, 105, 522, then there is 0<k<n/2 such that

k^2+(n-k)^2-3(n-1 mod 2) is prime and also JacobiSymbol[k,p]=1 for any prime divisor p of n+3(n-1 mod 2).

This conjecture has been verified for n up to 2*10^8. It is stronger than Ming-Zhi Zhang's conjecture that any odd integer n>1 can be written as x+y (x,y>0) with x^2+y^2 prime (see A036468).

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

EXAMPLE

a(33)=1 since 4^2+29^2=857 is prime and JacobiSymbol[4,33]=1.

a(24)=1 since 10^2+14^2-3=293 is prime and JacobiSymbol[10,24+3]=1.

MATHEMATICA

a[n_]:=a[n]=Sum[If[PrimeQ[k^2+(n-k)^2-3Mod[n-1, 2]]==True&&JacobiSymbol[k, n+3Mod[n-1, 2]]==1, 1, 0], {k, 1, (n-1)/2}]

Do[Print[n, " ", a[n]], {n, 1, 100}]

CROSSREFS

Cf. A036468, A191004, A185150, A220554.

Sequence in context: A254668 A306250 A145443 * A207385 A306242 A118144

Adjacent sequences:  A192053 A192054 A192055 * A192057 A192058 A192059

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Dec 30 2012

STATUS

approved

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Last modified November 14 12:16 EST 2019. Contains 329113 sequences. (Running on oeis4.)