A080909 a(n) = (2n+1)! modulo 4n+3, |a(n)| <= 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, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1

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

G. H. Hardy and E. M. Wright, An introduction to the theory of number, fourth edition, 1960, section 7.7: the residue of ((p-1)/2)!.

Links

Table of n, a(n) for n=0..82.

Formula

a(n) = mods((2*n+1)!, 4*n+3).

Example

a(3) = 0 since 7! == 0 (mod 15).
a(4) = 1 since 9! == -1 (mod 19).

Maple

seq(mods((2*n+1)!, 4*n+3), n=0..100);

Mathematica

a[ n_] := Mod[(2*n+1)!, 4*n+3, -1]; (* Michael Somos, Jul 25 2023 *)

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: A267533 A118274 A275737 * A087755 A050072 A267576

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

Name edited by Michael Somos, Jul 25 2023

Status

approved