The OEIS mourns the passing of Jim Simons and is grateful to the Simons Foundation for its support of research in many branches of science, including the OEIS.
login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A306530 a(n) is the smallest prime q such that Kronecker(q, prime(n)) = 1. 4
7, 7, 11, 2, 3, 3, 2, 5, 2, 5, 2, 3, 2, 11, 2, 7, 3, 3, 17, 2, 2, 2, 3, 2, 2, 5, 2, 3, 3, 2, 2, 3, 2, 5, 5, 2, 3, 41, 2, 13, 3, 3, 2, 2, 7, 2, 5, 2, 3, 3, 2, 2, 2, 3, 2, 2, 5, 2, 3, 2, 7, 17, 7, 2, 2, 7, 5, 2, 3, 3, 2, 2, 2, 3, 5, 2, 5, 3, 2, 2, 3, 3, 2, 2, 2, 3, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
For n >= 2, a(n) is the smallest prime quadratic residue modulo the n-th prime.
Also for n >= 2, a(n) is the smallest prime that decomposes in the quadratic field Q[sqrt((-1)^((p-1)/2)*p)], p = prime(n). Using this definition, a(1) should have been 5 because for p = 2, Q[sqrt((-1)^((p-1)/2)*p)] = Q[sqrt(2*i)] = Q[1+i] = Q[i], in which 5 decomposes.
For most n, a(n) is relatively small. Among [1, 10000], there are only 669 n's that violate a(n) < prime(n)/n and 97 n's > 1 that violate a(n) < prime(n)*log(log(n))/n. In fact, if we ignore the first three terms, the only terms among the first 10000 ones that seem unusually large are a(14) = 11, a(19) = 17, a(38) = 41, a(62) = 17, a(1137) = 29, a(1334) = 29, a(3935) = 37, a(7309) = 43, a(8783) = 37 and a(8916) = 41, with the corresponding primes 43, 67, 163, 293, 9173, 10987, 37123, 74093, 90787, 92333.
For every prime p there are infinitely many n such that a(n)=p. Indeed, using quadratic reciprocity, for each prime p_j <= p we can choose k_j coprime to p_j, such that p_j is a quadratic nonresidue (if p_j < p) or residue (if p_j = p) mod q for every prime q == k_j (mod p_j). Dirichlet's theorem on primes in arithmetic progressions implies there are infinitely many primes q with q == k_j (mod p_j) for all j. Then a(n) = p where q = prime(n). - Robert Israel, Mar 26 2019
a(n) is the smallest prime q such that the congruence x^2 == q (mod p) has a solution x, where p = prime(n). For n > 1, a(n) is the smallest prime q such that q^((p-1)/2) == 1 (mod p), where odd p = prime(n). - Thomas Ordowski, Apr 29 2019
LINKS
EXAMPLE
2, 3, 5, 7, ..., 37 are all quadratic nonresidues modulo prime(38) = 163, while 41 is a quadratic residue modulo 163, so a(38) = 41.
MAPLE
f:= proc(n) local q, p;
q:= ithprime(n);
p:= 1:
do
p:= nextprime(p);
if numtheory:-jacobi(p, q)=1 then return p fi
od;
end proc:
map(f, [$1..100]); # Robert Israel, Mar 26 2019
MATHEMATICA
a[n_] := Module[{i = 1}, While[KroneckerSymbol[Prime[i], Prime[n]] != 1, i++]; Prime[i]];
Array[a, 100] (* Jean-François Alcover, Jun 08 2020, after PARI *)
PROG
(PARI) a(n)=my(i=1); while(kronecker(prime(i), prime(n))!=1, i++); prime(i)
CROSSREFS
Cf. A053760 (smallest (prime) quadratic nonresidue modulo prime(n)).
Cf. A024704 (a(n)=2).
Sequence in context: A099290 A224895 A103569 * A094460 A070642 A064496
KEYWORD
nonn
AUTHOR
Jianing Song, Feb 22 2019
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified May 31 23:52 EDT 2024. Contains 373008 sequences. (Running on oeis4.)