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A163366
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(-1)^floor( (prime(i)+2)/2 ) mod prime(i).
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2
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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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 (jsondow(AT)alumni.princeton.edu), May 14 2010]
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REFERENCES
| Carl-Erik Froeberg, On sums and products of quadratic residues, BIT, Nord. Tidskr. Inf.-behandl. 11 (1971) 389-398. [Jonathan Sondow (jsondow(AT)alumni.princeton.edu), May 14 2010]
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LINKS
| Rahul Gupta, Algorithmic Number Theory, Section 24.5 [Jonathan Sondow (jsondow(AT)alumni.princeton.edu), May 14 2010]
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FORMULA
| a(n)*A177863(n) == -1 (mod prime(n)), by Wilson's theorem. [Jonathan Sondow (jsondow(AT)alumni.princeton.edu), May 14 2010]
a(n) = A177860(n) modulo prime(n). [Jonathan Sondow (jsondow(AT)alumni.princeton.edu), May 14 2010]
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EXAMPLE
| 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 (jsondow(AT)alumni.princeton.edu), May 14 2010]
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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 (jsondow(AT)alumni.princeton.edu), May 14 2010 *)
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CROSSREFS
| Cf. A152680, A005098, A002144, A009003.
Cf. A177860, A177863 [Jonathan Sondow (jsondow(AT)alumni.princeton.edu), May 14 2010]
Sequence in context: A111636 A146990 A051433 * A181145 A140070 A158815
Adjacent sequences: A163363 A163364 A163365 * A163367 A163368 A163369
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KEYWORD
| nonn
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AUTHOR
| Peter Luschny (peter(AT)luschny.de), Jul 25 2009
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