A349350 Dirichlet inverse of A057521, the powerful part of n.

1, -1, -1, -3, -1, 1, -1, -1, -8, 1, -1, 3, -1, 1, 1, 5, -1, 8, -1, 3, 1, 1, -1, 1, -24, 1, -10, 3, -1, -1, -1, 7, 1, 1, 1, 24, -1, 1, 1, 1, -1, -1, -1, 3, 8, 1, -1, -5, -48, 24, 1, 3, -1, 10, 1, 1, 1, 1, -1, -3, -1, 1, 8, -3, 1, -1, -1, 3, 1, -1, -1, 8, -1, 1, 24, 3, 1, -1, -1, -5, 28, 1, -1, -3, 1, 1, 1, 1, -1, -8

Offset: 1

Author

Antti Karttunen, Nov 18 2021

Keywords

• sign
• mult

Comments

Multiplicative because A057521 is.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A057521.

Cf. also A349340, A349442.

Programs

Mathematica

f[p_, e_] := Module[{B = 1 + p - 2*p^2, C = Sqrt[1 + 2*p - 3*p^2]}, FullSimplify[((B - C)*(p - 1 + C)^(e - 1) - (B + C)*(p - 1 - C)^(e - 1))/(2^e*C)]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Dec 24 2023 *)

PARI

A057521(n) = { my(f=factor(n)); prod(i=1, #f~, if(f[i, 2]>1, f[i, 1]^f[i, 2], 1)); }; \\ From A057521
memoA349350 = Map();
A349350(n) = if(1==n,1,my(v); if(mapisdefined(memoA349350,n,&v), v, v = -sumdiv(n,d,if(dA057521(n/d)*A349350(d),0)); mapput(memoA349350,n,v); (v)));

Formula

a(1) = 1; a(n) = -Sum_{d|n, d < n} A057521(n/d) * a(d).

Let p be a prime, B = 1 + p - 2*p^2 and C = sqrt(1 + 2*p - 3*p^2). Then the sequence is multiplicative with a(p^e) = ((B-C)*(p-1+C)^(e-1) - (B+C)*(p-1-C)^(e-1))/(2^e*C). - Sebastian Karlsson, Dec 02 2021