

A192056


a(n) = {0<k<n/2: k^2+(nk)^23(n1 mod 2) is prime and JacobiSymbol[k,n+3(n1 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
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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+(nk)^23(n1 mod 2) is prime and also JacobiSymbol[k,p]=1 for any prime divisor p of n+3(n1 mod 2).
This conjecture has been verified for n up to 2*10^8. It is stronger than MingZhi 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



EXAMPLE

a(33)=1 since 4^2+29^2=857 is prime and JacobiSymbol[4,33]=1.
a(24)=1 since 10^2+14^23=293 is prime and JacobiSymbol[10,24+3]=1.


MATHEMATICA

a[n_]:=a[n]=Sum[If[PrimeQ[k^2+(nk)^23Mod[n1, 2]]==True&&JacobiSymbol[k, n+3Mod[n1, 2]]==1, 1, 0], {k, 1, (n1)/2}]
Do[Print[n, " ", a[n]], {n, 1, 100}]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



