|
| |
|
|
A080909
|
|
(2n+1)! modulo 4n+3.
|
|
0
|
|
|
|
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
|
Table of n, a(n) for n=0..70.
|
|
|
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: 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
|
|
|
STATUS
|
approved
|
| |
|
|
