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A080909 (2n+1)! modulo 4n+3. +0
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 (list; graph; listen)
OFFSET

0,1

COMMENT

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)!

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;

CROSSREFS

Sequence in context: A011748 A130304 A118274 this_sequence A087755 A050072 A131309

Adjacent sequences: A080906 A080907 A080908 this_sequence A080910 A080911 A080912

KEYWORD

sign

AUTHOR

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

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Last modified August 19 23:53 EDT 2008. Contains 142930 sequences.


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