A163366 a(n) = (-1)^floor((prime(n)+2)/2) mod prime(n).

1, 1, 4, 1, 1, 12, 16, 1, 1, 28, 1, 36, 40, 1, 1, 52, 1, 60, 1, 1, 72, 1, 1, 88, 96, 100, 1, 1, 108, 112, 1, 1, 136, 1, 148, 1, 156, 1, 1, 172, 1, 180, 1, 192, 196, 1, 1, 1, 1, 228, 232, 1, 240, 1, 256, 1, 268, 1, 276, 280, 1, 292, 1, 1, 312, 316, 1, 336, 1, 348, 352, 1, 1, 372, 1

Offset: 1

Author

Peter Luschny, Jul 25 2009

Keywords

• nonn

Comments

Remove the '1's from the sequence to get A152680.

Product modulo p of the quadratic residues of p, where p = prime(n). [Jonathan Sondow, May 14 2010]

Examples

			a(4) = 1 because the quadratic residues of prime(4) = 7 are 1, 2, and 4, and 1*2*4 = 8 == 1 (mod 7). - _Jonathan Sondow_, May 14 2010
		

References

• Carl-Erik Froeberg, On sums and products of quadratic residues, BIT, Nord. Tidskr. Inf.-behandl. 11 (1971) 389-398. [Jonathan Sondow, May 14 2010]

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Rahul Gupta, Algorithmic Number Theory, Section 24.5 [_Jonathan Sondow_, May 14 2010]

Crossrefs

Cf. A152680, A005098, A002144, A009003.

Cf. A177860, A177863. - Jonathan Sondow, May 14 2010

Programs

Maple

seq((-1)^iquo(ithprime(i)+2,2) mod ithprime(i),i=1..113);

Mathematica

Table[Mod[ Apply[Times, Flatten[Position[ Table[JacobiSymbol[i, Prime[n]], {i, 1, Prime[n] - 1}], 1]]], Prime[n]], {n, 1, 80}] (* Jonathan Sondow, May 14 2010 *)

Formula

a(n)*A177863(n) == -1 (mod prime(n)), by Wilson's theorem. - Jonathan Sondow, May 14 2010

a(n) = A177860(n) modulo prime(n). - Jonathan Sondow, May 14 2010