The On-Line Encyclopedia of Integer Sequences (OEIS)

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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).
(history; published version)

#50 by Bruno Berselli at Thu Aug 29 11:39:23 EDT 2019

STATUS

reviewed

approved

#49 by Joerg Arndt at Thu Aug 29 10:53:37 EDT 2019

STATUS

proposed

reviewed

#48 by Jianing Song at Thu Aug 29 10:27:06 EDT 2019

STATUS

editing

proposed

#47 by Jianing Song at Thu Aug 29 10:27:00 EDT 2019

COMMENTS

For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 137). If k is a cube in 7-adic field, then k has exactly three cubic roots.

STATUS

proposed

editing

Discussion

Thu Aug 29

10:27

Jianing Song: Sorry.

#46 by Jianing Song at Thu Aug 29 02:57:28 EDT 2019

STATUS

editing

proposed

Discussion

Thu Aug 29

10:22

Joerg Arndt: Second comment: "mod 13"?

#45 by Jianing Song at Thu Aug 29 02:54:40 EDT 2019

CROSSREFS

this sequence, A319098, A319199 (7-adic, 76^(1/3));

#44 by Jianing Song at Thu Aug 29 02:53:24 EDT 2019

CROSSREFS

Cf. A319297, A319305, A319555.

#43 by Jianing Song at Thu Aug 29 02:52:25 EDT 2019

CROSSREFS

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, 7^(1/3));

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

#42 by Jianing Song at Thu Aug 29 02:43:51 EDT 2019

NAME

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

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 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.

LINKS

Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

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.

#41 by Jianing Song at Thu Aug 29 01:05:59 EDT 2019

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.