A342392 Cubefree numbers a > 1 such that the ring of integers of Q(a^(1/3)) is Z[a^(1/3)].

2, 3, 5, 6, 7, 11, 13, 14, 15, 21, 22, 23, 29, 30, 31, 33, 34, 38, 39, 41, 42, 43, 47, 51, 57, 58, 59, 61, 65, 66, 67, 69, 70, 74, 77, 78, 79, 83, 85, 86, 87, 93, 94, 95, 97, 101, 102, 103, 105, 106, 110, 111, 113, 114, 115, 119, 122, 123, 129, 130, 131, 133

Offset

1,1

Comments

Squarefree numbers not congruent to 1 or 8 modulo 9.

For k > 1, a != 1 being a squarefree number (a != -1 unless k is a power of 2), then the ring of integers of Q(a^(1/k)) is Z[a^(1/k)] if and only if: for every p dividing k, we have a^(p-1) !== 1 (mod p^2). In other words, O_Q(a^(1/k)) = Z[a^(1/k)] if and only if none of the prime factors of k is a Wieferich prime of base a. See Theorem 5.3 of the paper of Keith Conrad.

In general, if a^d == 1 (mod p^2) for some d|(p-1), then it is easy to show that x = (1 + a^(d/p) + a^(2*d/p) + ... + a^((p-1)*d/p))/p is an algebraic integer not in Z[a^(1/p)].

The asymptotic density of this sequence is 9/(2*Pi^2) = 0.455945... (A088245). - Amiram Eldar, Mar 11 2021

Links

Jianing Song, Table of n, a(n) for n = 1..15000

Keith Conrad, The ring of integers in a radical extension

Example

The ring of integers of Q(2^(1/3)) is {a + b*2^(1/3) + c*4^(1/3): a,b,c in Z} = Z[2^(1/3)], so 2 is a term.

Prog

(PARI) isA342392(n) = (n^2%9!=1) && issquarefree(n)

Crossrefs

Cf. A004709 (cubefree numbers), A088245, A342393.

Sequence in context: A080980 A334067 A134669 * A053328 A333786 A089633

Adjacent sequences: A342389 A342390 A342391 * A342393 A342394 A342395

Keyword

nonn,easy

Author

Jianing Song, Mar 10 2021

Status

approved