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%I #12 Sep 08 2022 08:45:30
%S 2,2,3,4,6,7,9,10,12,15,16,19,21,22,24,27,30,31,34,36,37,40,42,45,49,
%T 51,52,54,55,57,64,66,69,70,75,76,79,82,84,87,90,91,96,97,99,100,106,
%U 112,114,115,117,120,121,126,129,132,135,136,139,141,142,147,154,156,157
%N Number of quadratic residues (including 0) modulo the n-th prime.
%C The number of squares (quadratic residues including 0) modulo a prime p (sequence A096008 with every "1" prefixed by a "0") equals 1+floor(p/2), or ceiling(p/2) = (p+1)/2 if p is odd. (In fields of characteristic 2, all elements are squares.) See A130290(n)=A130291(n)-1 for number of nonzero residues. For all n>0, A130291(n+1) = A111333(n+1) = A006254(n) = A005097(n)-1 = A102781(n+1)-1 = A102781(n+1)-1 = A130290(n+1)-1.
%H Vincenzo Librandi, <a href="/A130291/b130291.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/QuadraticResidue.html">Quadratic Residue</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Quadratic_residue">Quadratic Residue</a>
%F a(n) = floor( A000040(n)/2 )+1
%e a(1)=2 since both elements of Z/2Z are squares.
%e a(3)=0 since 0=0^2, 1=1^2=(-1)^2 and 4=2^2=(-2)^2 are squares in Z/5Z.
%e a(1000000) = 7742932 = (p[1000000]+1)/2.
%t Quotient[Prime[Range[200]], 2] + 1 (* _Vincenzo Librandi_, Jan 16 2013 *)
%o (PARI) A130291(n) = 1+prime(n)>>1
%o (Magma) [Floor((NthPrime(n))/2)+1: n in [1..60]]; // _Vincenzo Librandi_, Jan 16 2013
%Y Essentially the same as A006254.
%Y Cf. A005097 (Odd primes - 1)/2, A102781 (Integer part of n#/(n-2)#/2#), A102781 (Number of even numbers less than the n-th prime), A063987 (quadratic residues modulo the n-th prime), A006254 (Numbers n such that 2n-1 is prime), A111333 (Number of odd numbers <= n-th prime), A000040 (prime numbers), A130290 (number of nonzero residues modulo primes).
%K easy,nonn
%O 1,1
%A _M. F. Hasler_, May 21 2007
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