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A080909
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(2n+1)! modulo 4n+3.
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0
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1, -1, -1, 0, -1, 1, 0, 1, 0, 0, -1, -1, 0, 0, 1, 0, -1, 1, 0, -1, 1, 0, 0, 0, 0
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| If 4n+3 is composite, then a(n)=0. If 4n+3 is prime, then a(n)=(-1)^m where m is the number of quadratic non-residues less than or equal to 2n+1. Is there a way to predict whether a(n)=1 or a(n)=-1 ?
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REFERENCES
| Hardy G. H., Wright E. M., An introduction to the theory of number (fourth edition, 1960), section 7.7: the residue of ((p-1)/2)!
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FORMULA
| a(n) = mods((2*n+1)!, 4*n+3)
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EXAMPLE
| a(3)=0 since 7! = 0 modulo 15 and a(4)=1 since 9! = -1 modulo 19.
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MAPLE
| for n from 0 to 20 do mods((2*n+1)!, 4*n+3) end do;
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CROSSREFS
| Sequence in context: A145361 A130304 A118274 * A087755 A050072 A156707
Adjacent sequences: A080906 A080907 A080908 * A080910 A080911 A080912
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KEYWORD
| sign
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AUTHOR
| Christophe Leuridan (ChristopheLeuridan(AT)ujf-grenoble.fr), Apr 01 2003
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