A347238 Dirichlet inverse of A347236.

1, -1, -2, -6, -2, 2, -4, 0, -15, 2, -2, 12, -4, 4, 4, 0, -2, 15, -4, 12, 8, 2, -6, 0, -35, 4, 0, 24, -2, -4, -6, 0, 4, 2, 8, 90, -4, 4, 8, 0, -2, -8, -4, 12, 30, 6, -6, 0, -77, 35, 4, 24, -6, 0, 4, 0, 8, 2, -2, -24, -6, 6, 60, 0, 8, -4, -4, 12, 12, -8, -2, 0, -6, 4, 70, 24, 8, -8, -4, 0, 0, 2, -6, -48, 4, 4, 4, 0, -8

Offset

1,3

Comments

Multiplicative because A347236 is.

It seems that A046099 gives the positions of zeros.

This follows from the formula for a(p^e). - Sebastian Karlsson, Sep 01 2021

Links

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

Index entries for primes, gaps between

Index entries for sequences computed from indices in prime factorization

Formula

a(1) = 1; a(n) = -Sum_{d|n, d < n} A347236(n/d) * a(d).

a(n) = A347239(n) - A347236(n).

For all n >= 1, a(A000040(n)) = -A001223(n).

Multiplicative with a(p^e) = p - A151800(p) if e = 1, -p*A151800(p) if e = 2 and 0 if e > 2. - Sebastian Karlsson, Sep 01 2021

Prog

(PARI)
up_to = 16384;
DirInverseCorrect(v) = { my(u=vector(#v)); u[1] = (1/v[1]); for(n=2, #v, u[n] = (-u[1]*sumdiv(n, d, if(d<n, v[n/d]*u[d], 0)))); (u) }; \\ Compute the Dirichlet inverse of the sequence given in input vector v.
A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
A061019(n) = (((-1)^bigomega(n))*n);
A347236(n) = sumdiv(n, d, A061019(d)*A003961(n/d));
v347238 = DirInverseCorrect(vector(up_to, n, A347236(n)));
A347238(n) = v347238[n];

Crossrefs

Cf. A000040, A001223, A003961, A046099, A061019, A347236, A347239.

Cf. A151800.

Sequence in context: A289382 A062539 A064136 * A171898 A330541 A320575

Adjacent sequences: A347235 A347236 A347237 * A347239 A347240 A347241

Keyword

sign,mult

Author

Antti Karttunen, Aug 24 2021

Status

approved