A349911 Dirichlet inverse of A336466, which is fully multiplicative with a(p) = oddpart(p-1).
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
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16384
Programs
-
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 *)
-
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
A336466(n/d)*A349911(d),0)); mapput(memoA349911,n,v); (v)));
Comments