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Dirichlet inverse of A057521, the powerful part of n.
3

%I #15 Dec 24 2023 02:41:44

%S 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,

%T -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,

%U -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

%N Dirichlet inverse of A057521, the powerful part of n.

%C Multiplicative because A057521 is.

%H Antti Karttunen, <a href="/A349350/b349350.txt">Table of n, a(n) for n = 1..20000</a>

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

%F 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

%t 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 *)

%o (PARI)

%o A057521(n) = { my(f=factor(n)); prod(i=1, #f~, if(f[i, 2]>1, f[i, 1]^f[i, 2], 1)); }; \\ From A057521

%o memoA349350 = Map();

%o A349350(n) = if(1==n,1,my(v); if(mapisdefined(memoA349350,n,&v), v, v = -sumdiv(n,d,if(d<n,A057521(n/d)*A349350(d),0)); mapput(memoA349350,n,v); (v)));

%Y Cf. A057521.

%Y Cf. also A349340, A349442.

%K sign,mult

%O 1,4

%A _Antti Karttunen_, Nov 18 2021