

A053760


Smallest positive quadratic nonresidue modulo p, where p is the nth prime.


13



2, 2, 2, 3, 2, 2, 3, 2, 5, 2, 3, 2, 3, 2, 5, 2, 2, 2, 2, 7, 5, 3, 2, 3, 5, 2, 3, 2, 2, 3, 3, 2, 3, 2, 2, 3, 2, 2, 5, 2, 2, 2, 7, 5, 2, 3, 2, 3, 2, 2, 3, 7, 7, 2, 3, 5, 2, 3, 2, 3, 2, 2, 2, 11, 5, 2, 2, 5, 2, 2, 3, 7, 3, 2, 2, 5, 2, 2, 3, 7, 2, 2, 7, 5, 3, 2, 3, 5, 2, 3, 2, 13, 3, 2, 2, 5, 2, 3, 2, 2, 2, 2, 2
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OFFSET

1,1


COMMENTS

Assuming the Generalized Riemann Hypothesis, Montgomery proved a(n) << (log p(n))^2, meaning that there is a constant c such that a(n) <= c*(log p(n))^2.  Jonathan Vos Post, Jan 06 2007
a(n) < 1 + sqrt(p), where p is the nth prime (Theorem 3.9 in Niven, Zuckerman, and Montgomery).  Jonathan Sondow, May 13 2010
Treviño proves that a(n) < 1.1 p^(1/4) log p for n > 2 where p is the nth prime.  Charles R Greathouse IV, Dec 06 2012
a(n) is always a prime, because if x*y is a nonresidue, then x or y must also be a nonresidue.  Jonathan Sondow, May 02 2013


REFERENCES

P. Erdős, Remarks on number theory. I., Mat. Lapok 12 (1961) 1017; Math. Rev. 26 #2410.
S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 9498.
H. L. Montgomery, Topics in Multiplicative Number Theory, 3rd ed., Lecture Notes in Mathematics, Vol. 227 (1971), MR 49:2616.
Ivan Niven, Herbert S. Zuckerman, and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991, p. 147.
P. Ribenboim, The New Book of Prime Number Records, 3rd ed., SpringerVerlag 1996; Math. Rev. 96k:11112.


LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000
R. Baillie and S. S. Wagstaff, Lucas pseudoprimes, Math. Comp. 35 (1980) 13911417; Math. Rev. 81j:10005.
S. R. Finch, Quadratic Residues
K. Matthews, Finding n(p), the least quadratic nonresidue (mod p)
Enrique Treviño, The least kth power nonresidue, 2011 preprint
Eric Weisstein's World of Mathematics, Quadratic Nonresidue


EXAMPLE

The 5th prime is 11, and the positive quadratic residues mod 11 are 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 5 and 5^2 = 3. Since 2 is missing, a(5) = 2.
The only positive quadratic redidue mod 2 is 1, so a(1)=2.


MATHEMATICA

Table[ p = Prime[n]; First[ Select[ Range[p], JacobiSymbol[#, p] != 1 &]], {n, 1, 100}] (* Jonathan Sondow, Mar 03 2013 *)


PROG

(PARI) residue(n, m)={local(r); r=0; for(i=0, floor(m/2), if(i^2%m==n, r=1)); r
A053760(n)={local(r, m); r=0; m=0; while(r==0, m=m+1; if(!residue(m, prime(n)), r=1)); m} \\ Michael B. Porter, May 02 2010
(PARI) qnr(p)=my(m); while(1, if(!issquare(Mod(m++, p)), return(m)))
a(n)=if(n>1, qnr(prime(n)), 2) \\ Charles R Greathouse IV, Feb 27 2013


CROSSREFS

Cf. A000229.
Sequence in context: A219252 A085694 A160493 * A223942 A129654 A138789
Adjacent sequences: A053757 A053758 A053759 * A053761 A053762 A053763


KEYWORD

nonn


AUTHOR

Steven Finch, Apr 05 2000


EXTENSIONS

More terms from James A. Sellers, Apr 08 2000


STATUS

approved



