A373371 a(n) = 1 if the sum of prime factors with repetition is a multiple of 3, otherwise 0.

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

Offset

1

Comments

a(n) = 1 if the multiplicities of prime factors of the forms 3m+1 (A002476) and 3m-1 (A003627) are equal modulo 3, otherwise 0. - Antti Karttunen, Jun 13 2024

Links

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

Index entries for characteristic functions

Formula

a(n) = [A001414(n) == 0 (mod 3)], where [ ] is the Iverson bracket.

From Antti Karttunen, Jun 13 2024: (Start)

a(n) = [A373591(n) == A373592(n) (mod 3)].

a(n) = a(n/A038500(n)) = A359430(n/A038500(n)) = A369658(n/A038500(n)).

(End)

Prog

(PARI)
A001414(n) = ((n=factor(n))[, 1]~*n[, 2]); \\ From A001414.
A373371(n) = !(A001414(n)%3);
(PARI) A373371(n) = { my(f = factor(n), m1=0, m2=0); for(i = 1, #f~, if(1==(f[i, 1]%3), m1 += f[i, 2], if(2==(f[i, 1]%3), m2 += f[i, 2]))); 0==((m1-m2)%3); }; \\ Antti Karttunen, Jun 13 2024

Crossrefs

Characteristic function of A289142.

Cf. A001414, A002476, A003627, A038500, A373591, A373592.

Cf. also A359430, A369658, A373372.

Sequence in context: A190236 A190224 A352678 * A321692 A355943 A102242

Adjacent sequences: A373368 A373369 A373370 * A373372 A373373 A373374

Keyword

nonn,changed

Author

Antti Karttunen, Jun 02 2024

Status

approved