A319199 One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 6 (mod 7) case (except for n = 0).

0, 6, 34, 83, 769, 3170, 36784, 36784, 3330956, 26390160, 187804588, 470279837, 470279837, 83518003043, 180407013450, 180407013450, 23918214563165, 90384075702367, 1020906131651195, 7534560523292991, 53130141264785563, 212714673860009565, 1888352266109861586

Offset: 0

Author

Jianing Song, Aug 27 2019

Keywords

• nonn

Comments

For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 6 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.

Examples

			The unique number k in [1, 7^2] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 34, so a(2) = 24.
The unique number k in [1, 7^3] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 83, so a(3) = 122.
		

Links

• Wikipedia, p-adic number

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));

A319097, A319098, this sequence (7-adic, 6^(1/3));

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

Programs

PARI

a(n) = lift(sqrtn(6+O(7^n), 3))

Formula

a(n) = A319097(n)*(A210852(n)-1) mod 7^n = A319097(n)*A210852(n)^2 mod 7^n.

a(n) = A319098(n)*(A212153(n)-1) mod 7^n = A319098(n)*A212153(n)^2 mod 7^n.