OFFSET
1,1
COMMENTS
Write P = 608981813017, then Sum_{q prime<=p} kronecker(-3,q) <= -1 for 2 <= p < P (see A096449), and is equal to 0 for p = P.
For n = 3^e*m^2, e odd, gcd(m,3) = 1:
(a) if m has no prime factor < P, then Sum_{q prime<=p} kronecker(-n,q) = Sum_{q prime<=p} kronecker(-3,q) < 0 for primes p < P, so a(n) > P.
(b) if m has prime factors < P, let r be the largest such factor. Then
- Sum_{q prime<=p} kronecker(-n,q) = Sum_{q prime<=p} kronecker(-3,q) < 0 for primes p < lpf(m), where lpf = least prime factor;
- Sum_{q prime<=p} kronecker(-n,q) = Sum_{q prime<=p} kronecker(-3,q) - Sum_{q|m, q<P} kronecker(-3,q) for primes r <= p < P.
As a result, if Sum_{q|m, q<P} kronecker(-3,q) >= -1, then either lpf(m) < a(n) < r, or a(n) >= P.
Similarly, write P = 24996190781. For n = 1605^e*m^2, e odd, gcd(m,1605) = 1, if Sum_{q|m, q<P} kronecker(-1605,q) >= -9, then we have either lpf(m) < a(n) < r (where r is the largest prime factor < P of m) or a(n) >= P, unless the prime factors < P of m are exactly
- {2,7} (then a(n) = 11);
- {2,13,17,19,23,29} (then a(n) = 31);
- {7,13,17,19,23,29} (then a(n) = 31);
- {2,7,11,13,17,19,23,29} (then a(n) = 31).
See my link below for more details.
LINKS
Jianing Song, Table of n, a(n) for n = 1..10000
Jianing Song, Finding a(1605^e*m^2), e odd, gcd(m,1605) = 1
Wikipedia, Chebyshev's bias
PROG
(PARI) a(n) = {
if(issquare(n/3), my(P=608981813017, temp=n, f, s=0, maxp=0); while(temp%3==0, temp/=3); f=factor(temp); for(i=1, #f~, if(f[i, 1]<P, s+=kronecker(-3, f[i, 1]); maxp=f[i, 1], break())); if(s>=-1, my(i=0); forprime(p=2, maxp, i+=kronecker(-n, p); if(i>0, return(p))); i=-1-s; forprime(p=P, oo, i+=kronecker(-n, p); if(i>0, return(p))); ); ); \\ the case n = 3^e*m^2, Sum_{q|m, q<P} kronecker(-3, q) >= -1
my(i=0); forprime(p=2, oo, i+=kronecker(-n, p); if(i>0, return(p)))}
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Jan 06 2026
STATUS
approved
