|
| |
|
|
A342393
|
|
Cubefree numbers a > 1 such that the ring of integers of Q(a^(1/3)) is not Z[a^(1/3)].
|
|
2
|
|
|
|
4, 9, 10, 12, 17, 18, 19, 20, 25, 26, 28, 35, 36, 37, 44, 45, 46, 49, 50, 52, 53, 55, 60, 62, 63, 68, 71, 73, 75, 76, 82, 84, 89, 90, 91, 92, 98, 99, 100, 107, 109, 116, 117, 118, 121, 124, 126, 127, 132, 134, 140, 143, 145, 147, 148, 150, 153, 154, 156, 161
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Cubefree numbers that are not squarefree or are 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)].
There is no consecutive run of 8 numbers in this sequence since every 8 consecutive numbers must contain a multiple of 8, which is not cubefree. The first consecutive run of 7 numbers in this sequence is 59921 ~ 59927: 59922, 59924, 59925, 59926 and 59927 are not squarefree, 59921 == 8 (mod 9) and 59923 == 1 (mod 9).
The asymptotic density of this sequence is 1/zeta(3) - 9/(2*Pi^2) = 0.375962... - Amiram Eldar, Mar 11 2021
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
The ring of integers of Q(4^(1/3)) = Q(2^(1/3)) is not Z[4^(1/3)] but Z[2^(1/3)], so 4 is a term.
The ring of integers of Q(12^(1/3)) = Q(18^(1/3)) is not Z[12^(1/3)] nor Z[18^(1/3)] but Z[12^(1/3)+18^(1/3)], so both 12 and 18 are terms. Note that if x = 12^(1/3) + 18^(1/3), then 12^(1/3) = x^2 - 2*x - 12, 18^(1/3) = -x^2 + 3*x + 12, and the minimal polynomial of x is x^3 - 18*x - 30.
For a = 9c+1, x = (1 + a^(1/3) + a^(2/3))/3 satisfies the monic cubic x^3 - x^2 - 3c*x - 3c^2 = 0; For a = 9c-1, x = (1 - a^(1/3) + a^(2/3))/3 satisfies the monic cubic x^3 - x^2 + 3c*x - 3c^2 = 0. In both cases, x is an algebraic integer not in Z[a^(1/3)].
|
|
|
PROG
|
(PARI) isA342393(n) = if(n>1, my(e=vecmax(factor(n)[, 2]~)); if(e<3, e==2 || (n^2%9==1)), 0)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
