|
|
A319097
|
|
One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 3 (mod 7) case (except for n = 0).
|
|
11
|
|
|
0, 3, 24, 122, 808, 10412, 111254, 817148, 1640691, 24699895, 186114323, 186114323, 6118094552, 6118094552, 490563146587, 2525232365134, 26263039914849, 59495970484450, 1222648540420485, 6107889334151832, 74501260446390690, 234085793041614692, 1351177521208182706, 24810103812706111000, 134285093173029776372
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 3 mod 7 such that k^3 - 6 is divisible by 7^n.
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
The unique number k in [1, 7^2] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 24, so a(2) = 24.
The unique number k in [1, 7^3] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 122, so a(3) = 122.
|
|
PROG
|
(PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1+sqrt(-3+O(7^n)))/2)
|
|
CROSSREFS
|
Approximations of p-adic cubic roots:
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|