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A359155
Dirichlet inverse of A359154, where A359154 is multiplicative with a(p^e) = (-1)^(p*e).
3
1, -1, 1, 0, 1, -1, 1, 0, 0, -1, 1, 0, 1, -1, 1, 0, 1, 0, 1, 0, 1, -1, 1, 0, 0, -1, 0, 0, 1, -1, 1, 0, 1, -1, 1, 0, 1, -1, 1, 0, 1, -1, 1, 0, 0, -1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, -1, 1, 0, 1, -1, 0, 0, 1, -1, 1, 0, 1, -1, 1, 0, 1, -1, 0, 0, 1, -1, 1, 0, 0, -1, 1, 0, 1, -1, 1, 0, 1, 0, 1, 0, 1, -1, 1, 0, 1, 0, 0, 0, 1, -1, 1, 0, 1
OFFSET
1
LINKS
FORMULA
Multiplicative with a(p) = (-1)^(1+p), and a(p^e) = 0 if e > 1.
a(n) = A008683(n) * A359154(n).
a(1) = 1, and for n > 1, a(n) = -Sum_{d|n, d<n} A359154(n/d) * a(d).
For all n >= 1, a(A003961(n)) = A008966(n).
Dirichlet g.f.: (zeta(s)/zeta(2*s))*((2^s-1)/(2^s+1)). - Amiram Eldar, Dec 29 2022
MATHEMATICA
f[p_, e_] := If[e == 1, 1, 0]; f[2, e_] := If[e == 1, -1, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Dec 29 2022 *)
PROG
(PARI) A359155(n) = { my(f = factor(n)); prod(k=1, #f~, (1==f[k, 2])*((-1)^(1+f[k, 1]))); };
(PARI) A359155(n) = { my(f = factor(n)); moebius(n)*prod(k=1, #f~, (-1)^(f[k, 1]*f[k, 2])); };
(Python)
from functools import reduce
from operator import ixor
from sympy import factorint
def A359155(n): return 0 if max((f:=factorint(n)).values(), default=0) > 1 else -1 if reduce(ixor, (p&1^1 for p in f.keys()), 0) else 1 # Chai Wah Wu, Dec 21 2022
CROSSREFS
KEYWORD
sign,mult
AUTHOR
Antti Karttunen, Dec 19 2022
STATUS
approved