OFFSET
1,1
COMMENTS
Analog of Moebius mu with sqrt(2) adjoined. Same as mu (A008683) except: 0 for even n (A005843) due to square (extended prime) factor (sqrt(2))^2 and rational primes of the form 8k+/-1 (A001132) factor into conjugate (extended prime) pairs (a + b*sqrt(2))(a - b*sqrt(2)), thus contributing +1 to the product instead of -1; e.g., 7 = (3 + sqrt(2))(3 - sqrt(2)).
For even n a(n) must be 0 because 2 is a square in the quadratic field and so the mu-analog is 0. Of course this coincidentally matches the 0's at even n in A087003. For odd n, from its definition as a product, |a(n)| MUST be the same as that of |mu(n)|. Since from the above we know that A087003(n) is the same as mu(n) at odd n, we can conclude that |a(n)| = |A087003(n)| for all n.
REFERENCES
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, Theorem 256, p. 221.
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..20000
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..100000
FORMULA
a(n) = 0 if n even or has a square prime factor, otherwise Product_{p divides n} (2 - |p mod 8|) where the product is taken over the primes.
From Jianing Song, Aug 30 2018: (Start)
Multiplicative with a(p^e) = 0 if p = 2 or e > 1, a(p) = 1 if p == +-1 (mod 8) and -1 if p == +-3 (mod 8).
For squarefree n, a(n) = Kronecker symbol (n, 2) (or (2, n)) = A091337(n). Also for these n, a(n) = A318608(n) if n even or n == 1 (mod 8), otherwise -A318608(n).
(End)
EXAMPLE
a(21) = (-1)*(+1) = -1 because 21 = 3*7 where 3 and 7 are congruent to +3 and -1 mod 8 respectively.
PROG
(PARI)
a(n)=
{
my(r=1, f=factor(n));
for(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]);
if(p==2||e>=2, r=0);
if((Mod(p, 8)==3||Mod(p, 8)==5)&e==1, r*=-1);
);
return(r);
} \\ Jianing Song, Aug 30 2018
CROSSREFS
KEYWORD
mult,easy,sign
AUTHOR
Marc LeBrun, Dec 17 2003
STATUS
approved