A353350 a(n) = 1 if A048675(n) is a multiple of 3, otherwise 0.

1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1

Offset

1

Comments

Proof for the formula a(n) = A353269(A332462(n)): This follows from a more general identity A048675(n) == A156552(A019565(A156552(n))) (mod 3). Applying A156552's inverse, the Doudna-map x -> A005940(1+x), to the both sides from the right we obtain A087808(n) == A048678(n) (mod 3), and from the respective recurrences of those two sequences it is easy to see that they are equal when reduced modulo 3.

Links

Antti Karttunen, Table of n, a(n) for n = 1..65537

Index entries for characteristic functions

Index entries for sequences computed from indices in prime factorization

Formula

a(n) = A079978(A048675(n)).

a(n) = A353269(A332462(n)). [See comments for a proof]

a(n) = 1 iff A332813(n) = 0, or equally iff A332823(n) = 0.

a(p) = 0 for all primes p.

a(n) = a(A003961(n)) = a(A348717(n)), for all n >= 1.

Mathematica

f[p_, e_] := e*2^(PrimePi[p] - 1); a[1] = 1; a[n_] := Boole @ Divisible[Plus @@ f @@@ FactorInteger[n], 3]; Array[a, 100] (* Amiram Eldar, Apr 15 2022 *)

Prog

(PARI)
A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
A353350(n) = (0==(A048675(n)%3));

Crossrefs

Characteristic function of A332820.

Cf. A003961, A048675, A079978, A332462, A332813, A332823, A348717,

Cf. A353348 (Dirichlet inverse), A353349 (sum with it), A353352 (inverse Möbius transform), A353353.

Cf. also A353269, A353380.

Sequence in context: A173864 A353348 A359826 * A354097 A374051 A354927

Adjacent sequences: A353347 A353348 A353349 * A353351 A353352 A353353

Keyword

nonn

Author

Antti Karttunen, Apr 15 2022

Status

approved