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A349911
Dirichlet inverse of A336466, which is fully multiplicative with a(p) = oddpart(p-1).
2
1, -1, -1, 0, -1, 1, -3, 0, 0, 1, -5, 0, -3, 3, 1, 0, -1, 0, -9, 0, 3, 5, -11, 0, 0, 3, 0, 0, -7, -1, -15, 0, 5, 1, 3, 0, -9, 9, 3, 0, -5, -3, -21, 0, 0, 11, -23, 0, 0, 0, 1, 0, -13, 0, 5, 0, 9, 7, -29, 0, -15, 15, 0, 0, 3, -5, -33, 0, 11, -3, -35, 0, -9, 9, 0, 0, 15, -3, -39, 0, 0, 5, -41, 0, 1, 21, 7, 0, -11, 0
OFFSET
1,7
COMMENTS
Multiplicative because A336466 is.
LINKS
FORMULA
a(1) = 1; a(n) = -Sum_{d|n, d < n} A336466(n/d) * a(d).
a(n) = A349912(n) - A336466(n).
MATHEMATICA
f[p_, e_] := ((p-1)/2^IntegerExponent[p-1, 2])^e; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; a[1] = 1; a[n_] := a[n] = -DivisorSum[n, a[#] * s[n/#] &, # < n &]; Array[a, 100] (* Amiram Eldar, Dec 08 2021 *)
PROG
(PARI)
A000265(n) = (n>>valuation(n, 2));
A336466(n) = { my(f=factor(n)); prod(k=1, #f~, A000265(f[k, 1]-1)^f[k, 2]); };
memoA349911 = Map();
A349911(n) = if(1==n, 1, my(v); if(mapisdefined(memoA349911, n, &v), v, v = -sumdiv(n, d, if(d<n, A336466(n/d)*A349911(d), 0)); mapput(memoA349911, n, v); (v)));
CROSSREFS
Cf. also A097945.
Sequence in context: A325846 A325735 A235794 * A090030 A293616 A211649
KEYWORD
sign,mult
AUTHOR
Antti Karttunen, Dec 08 2021
STATUS
approved