A279119 Lexicographically earliest sequence such that, for any distinct i and j, a(i)=a(j) implies gcd(i, j)=1.

0, 0, 0, 1, 0, 2, 0, 3, 1, 4, 0, 5, 0, 6, 3, 7, 0, 8, 0, 9, 4, 10, 0, 11, 1, 12, 6, 13, 0, 14, 0, 15, 7, 16, 2, 17, 0, 18, 9, 19, 0, 20, 0, 21, 10, 22, 0, 23, 1, 24, 12, 25, 0, 26, 5, 27, 13, 28, 0, 29, 0, 30, 15, 31, 6, 32, 0, 33, 16, 34, 0, 35, 0, 36, 18, 37

Offset

1,6

Comments

Also, for n>1, a(n) equals the index of the class of n relatively to the algorithm described in A275246 (i.e., if a(n)=k, then n is of class P_k).

From Rémy Sigrist, Dec 21 2016: (Start)

For any prime p, the sequence b_p(n)=a(p*n) is a bijection from A000027 to A001477:

- b_p is injective: b_p(n)=b_p(m) implies p*n=p*m or gcd(p*n,p*m)=1; as p>1, gcd(p*n,p*m)>1, so p*n=p*m and n=m.

- b_p is surjective: by contradiction: let k be the least number such that b_p(n) never equals k; we have a set of k terms (i_1,...,i_k) such that b_p(i_j) = j-1 for any j between 1 and k; let l be the least value such that p^l > max({1, i_1,...,i_k}). Then, by definition of a, a(p^l)=k, and b_p(p^(l-1))=k, which is a contraction.

(End)

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Formula

a(2*n) = n-1 for any n>0.

a(n)=0 iff n belongs to A008578.

a(n)=1 iff n belongs to A001248.

a(n)=2 iff n belongs to A089581.

a(n)=3 iff n belongs to A275246.

a(n)=4 iff n belongs to A275248.

a(n)=5 iff n belongs to A275249.

a(n)=6 iff n belongs to A275251.

a(n)=7 iff n belongs to A275252.

a(n)=8 iff n belongs to A275253.

Prog

(PARI) g = vector(76, i, 1); for (n=1, #g, a = 0; while (gcd(g[a+1], n)>1, a++); g[a+1] *= n; print1 (a ", "))

Crossrefs

Cf. A008578, A001248, A089581, A275246, A275248, A275249, A275251, A275252, A275253.

Sequence in context: A355526 A243056 A338568 * A249738 A364020 A363898

Adjacent sequences: A279116 A279117 A279118 * A279120 A279121 A279122

Keyword

nonn

Author

Rémy Sigrist, Dec 06 2016

Status

approved