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