cp's OEIS Frontend

A048280 Length of longest run of consecutive quadratic residues mod prime(n).

%I A048280 #41 May 23 2020 22:45:24
%S A048280 2,2,3,3,3,3,5,4,5,4,4,4,5,5,5,3,5,5,6,7,9,6,7,5,9,7,7,6,5,5,7,8,6,5,
%T A048280 4,7,6,6,6,6,6,6,7,9,7,6,7,7,7,5,6,7,13,7,6,7,8,7,10,6,9,9,7,11,9,5,8,
%U A048280 9,8,6,6,8,9,6,8,8,8,5,7,13,8,7,7,9,10,8,8,9,8,8,11,13,8,8,10,8,9,8,10,7,9,9,10,10,7,9
%N A048280 Length of longest run of consecutive quadratic residues mod prime(n).
%C A048280 0 and 1 are consecutive quadratic residues for any prime, so a(n) >= 2.
%C A048280 "Consecutive" allows wrap-around, so p-1 and 0 are consecutive. - _Robert Israel_, Jul 20 2014
%C A048280 A002307(n) is defined similarly, except that only positive reduced quadratic residues are counted. - _Jonathan Sondow_, Jul 20 2014
%C A048280 For longest runs of quadratic nonresidues, see A002308. - _Jonathan Sondow_, Jul 20 2014
%H A048280 Robert Israel, <a href="/A048280/b048280.txt">Table of n, a(n) for n = 1..10000</a>
%H A048280 P. Pollack and E. Treviño, <a href="http://pollack.uga.edu/mullin.pdf">The primes that Euclid forgot</a>, Amer. Math. Monthly, 121 (2014), 433-437.
%H A048280 Enrique Treviño, <a href="http://campus.lakeforest.edu/trevino/CorrigendumConstantCharacter.pdf">Corrigendum to “On the maximum number of consecutive integers on which a character is constant”</a>, Mosc. J. Comb. Number Theory 7 (2017), no. 3, 1-2.
%F A048280 a(n) < 2*sqrt(prime(n)) for n >= 1 (see Pollack and Treviño for n > 1). - _Jonathan Sondow_, Jul 20 2014
%F A048280 a(n) >= A002307(n). - _Jonathan Sondow_, Jul 20 2014
%F A048280 a(n) < 7 prime(n)^(1/4)log(prime(n)) for all n > 1, or a(n) < 3.2 prime(n)^(1/4)log(prime(n)) for n >= 10^13. - _Enrique Treviño_, Apr 16 2020
%e A048280 For n = 7, prime(7) = 17 has consecutive quadratic residues 15,16,0,1,2, and no longer sequence of consecutive quadratic residues, so a(7)=5.
%p A048280 A:= proc(n) local P, res, nonres, nnr;
%p A048280      P:= ithprime(n);
%p A048280      res:= {seq(i^2,i=0..floor((P-1)/2)};
%p A048280      nonres:= {$1..P-1} minus res;
%p A048280      nnr:= nops(nonres);
%p A048280      max(seq(nonres[i+1]-nonres[i]-1,i=1..nnr-1),nonres[1]-nonres[-1]+P-1)
%p A048280 end proc;
%p A048280 A(1):= 2:
%p A048280 seq(A(n),n=1..100); # _Robert Israel_, Jul 20 2014
%Y A048280 Cf. A002307, A048281.
%K A048280 nonn
%O A048280 1,1
%A A048280 _David W. Wilson_
%E A048280 Offset corrected to 1 and definition clarified by _Jonathan Sondow_ Jul 20 2014