1,3

Row lengths for formatting A063987 as a table: The number of nonzero quadratic residues modulo a prime p equals floor(p/2), or (p-1)/2 if p is odd. The number of squares including 0 is (p+1)/2, if p is odd (rows prime[i] of A096008 formatted as a table). In fields of characteristic 2, all elements are squares. For any m>0, floor(m/2) is the number of even positive integers less than or equal to m, so a(n) also equals the number of even positive integers less than or equal to the n-th prime. For all n>0, A130290(n+1) = A005097(n) = A102781(n+1) = A111332(n+1) = A130291(n+1)-1 = A111333(n+1)-1 = A006254(n)-1.

Eric Weisstein's World of Mathematics, Quadratic Residue

Wikipedia, Quadratic Residue

a(n) = floor( A000040(n)/2 ) = #{ even positive integers <= A000040(n) }

a(1)=1 since the only nonzero element of Z/2Z equals its square.

a(3)=2 since 1=1^2=(-1)^2 and 4=2^2=(-2)^2 are the only nonzero squares in Z/5Z.

a(1000000) = 7742931 = (p[1000000]-1)/2.

Quotient[Prime[Range[66]], 2] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 20 2008]

(PARI) A130290(n) = prime(n)>>1

Essentially the same as A005097.

Cf. A005097 (Odd primes - 1)/2, A102781 (Integer part of n#/(n-2)#/2#), A111332 (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), A130291.

Adjacent sequences: A130287 A130288 A130289 this_sequence A130291 A130292 A130293

Sequence in context: A093689 A097702 A082583 this_sequence A102781 A005097 A111332

easy,nonn

M. F. Hasler (Maximilian.Hasler(AT)gmail.com), May 21 2007