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).

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

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

Table of n, a(n) for n=0..24.

Wikipedia, p-adic number

Formula

a(n) = A319098(n)*(A210852(n)-1) mod 7^n = A319098(n)*A210852(n)^2 mod 7^n.

a(n) = A319199(n)*(A212153(n)-1) mod 7^n = A319199(n)*A212153(n)^2 mod 7^n.

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

Cf. A319297, A319305, A319555.

Approximations of p-adic cubic roots:

A290567 (5-adic, 2^(1/3));

A290568 (5-adic, 3^(1/3));

A309444 (5-adic, 4^(1/3));

this sequence, A319098, A319199 (7-adic, 6^(1/3));

A320914, A320915, A321105 (13-adic, 5^(1/3)).

Sequence in context: A151883 A009134 A009137 * A326789 A305543 A356363

Adjacent sequences: A319094 A319095 A319096 * A319098 A319099 A319100

Keyword

nonn

Author

Jianing Song, Aug 27 2019

Status

approved