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 A080909 (2n+1)! modulo 4n+3. 1
 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, -1, 1, 0, 0, 0, 0, -1, -1, 0, 1, 0, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, 0, -1, 0, -1, 0, 0, 1, 0, 0, 1, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 1, 0, 0, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 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 ? 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)! LINKS FORMULA a(n) = mods((2*n+1)!, 4*n+3) EXAMPLE a(3)=0 since 7! = 0 modulo 15 and a(4)=1 since 9! = -1 modulo 19. MAPLE for n from 0 to 20 do mods((2*n+1)!, 4*n+3) end do; PROG (PARI) a(n)= {v =(2*n+1)! % (4*n+3); if (2*v > 4*n+3, v -= 4*n+3); return (v); } \\ Michel Marcus, Jul 21 2013 CROSSREFS Sequence in context: A145361 A130304 A118274 * A087755 A050072 A156707 Adjacent sequences:  A080906 A080907 A080908 * A080910 A080911 A080912 KEYWORD sign AUTHOR Christophe Leuridan (ChristopheLeuridan(AT)ujf-grenoble.fr), Apr 01 2003 EXTENSIONS More terms from Michel Marcus, Jul 21 2013 STATUS approved

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