A080908 a(n) = the sign of r(n), where r(n) is the integer in [ -2n,2n] which is congruent to (2n)! modulo 4n+1

0, 1, -1, 1, -1, 0, 0, 1, 0, -1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, -1

Offset

0,1

Comments

If 4n+1 is composite, then a(n)=0, except when n=2. If 4n+1 is a prime number, then (2n)! is a square root of -1 modulo 4n+1 and a(n)=1 or a(n)=-1. Is there a simple way to predict whether a(n)=1 or a(n)=-1 ? The Maple program could be simplified by setting sign(0)=0, but I do not know how to do that.

References

Hardy, G. H. and Wright, E. M., 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..25.

Example

a(2) = -1 because 4! = 24 = -3 modulo 9 and a(5) = 0 because 10! = 0 modulo 21.

Maple

for n from 0 to 100 do (sign(2*mods((2*n)!, 4*n+1)+1) + sign(2*mods((2*n)!, 4*n+1)-1))/2 end do;

Crossrefs

Sequence in context: A099859 A176416 A102460 * A131720 A131719 A342460

Adjacent sequences: A080905 A080906 A080907 * A080909 A080910 A080911

Keyword

sign

Author

Christophe Leuridan (Christophe.Leuridan(AT)ujf-grenoble.fr), Apr 01 2003

Status

approved