OFFSET
1,1
COMMENTS
From Jianing Song, Jan 06 2026: (Start)
Write P = 2082927019, then Sum_{q prime<=p} kronecker(5,q) <= -2 for 2 < p < P, and is equal to -1 for p = P.
For n = 5^e*m^2, e odd, gcd(m,5) = 1:
(a) if m has no prime factor < P, then Sum_{q prime<=p} kronecker(n,q) = Sum_{q prime<=p} kronecker(5,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(5,q) < 0 for primes p < lpf(m), where lpf = least prime factor;
- Sum_{q prime<=p} kronecker(n,q) = Sum_{q prime<=p} kronecker(5,q) - Sum_{q|m, q<P} kronecker(5,q) for primes r <= p < P.
As a result, if Sum_{q|m, q<P} kronecker(5,q) >= -2, then either lpf(m) < a(n) < r, or a(n) >= P.
(End)
LINKS
Eric Chen, Table of n, a(n) for n = 1..4096
Wikipedia, Chebyshev's bias
FORMULA
PROG
(PARI) a(n) = my(i=0); forprime(p=2, oo, i+=kronecker(n, p); if(i>0, return(p))) \\ after Jianing Song in A306499
(PARI) a(n) = {
if(issquare(n/5), my(P=2082927019, temp=n, f, s=0, maxp=0); while(temp%5==0, temp/=5); f=factor(temp); for(i=1, #f~, if(f[i, 1]<P, s+=kronecker(5, f[i, 1]); maxp=f[i, 1], break())); if(s>=-2, my(i=0); forprime(p=2, maxp, i+=kronecker(n, p); if(i>0, return(p))); i=-2-s; forprime(p=P, oo, i+=kronecker(n, p); if(i>0, return(p))); ); ); \\ the case n = 5^e*m^2, Sum_{q|m, q<P} kronecker(5, q) >= -2
my(i=0); forprime(p=2, oo, i+=kronecker(n, p); if(i>0, return(p)))} \\ Jianing Song, Jan 06 2026; Should be a bit faster to generate a(n) for n = 5^e*m^2
CROSSREFS
KEYWORD
nonn
AUTHOR
Eric Chen, Jul 15 2019
STATUS
approved
