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A359835

a(n) = 0 if A353418(n) = 0, otherwise 1. Here A353418 is Dirichlet inverse of the characteristic function for numbers k at which points A156552(k) is a multiple of 3.

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

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OFFSET

COMMENTS

In contrast to A359836, which seems to satisfy A359836(n) <= A353269(n), here a(n) - A353269(n) may obtain any of the values -1, 0 or +1.

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..100000

Index entries for sequences computed from indices in prime factorization

FORMULA

a(n) = [A353418(n) != 0], where [ ] is the Iverson bracket.

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

a(n) >= A359836(n).

PROG