A322090 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 7 (mod 13) case (except for n = 0).

0, 7, 124, 124, 13306, 70428, 1926893, 40541365, 542529501, 2989721664, 45407719156, 458983194703, 18380587135073, 111572927624997, 2231698673770768, 2231698673770768, 462904735800587581, 5120821000082846468, 74324148355133549932, 1423789031778622267480, 10195310774031298931542

Offset

0,2

Comments

For n > 0, a(n) is the unique solution to x^2 == -3 (mod 13^n) in the range [0, 13^n - 1] and congruent to 7 modulo 13.

A322089 is the approximation (congruent to 6 mod 13) of another square root of -3 over the 13-adic field.

Links

Table of n, a(n) for n=0..20.

Wikipedia, p-adic number

Formula

For n > 0, a(n) = 13^n - A322089(n).

a(n) = Sum_{i=0..n-1} A322092(i)*13^i.

a(n) = A286840(n)*A322085(n) mod 13^n = A286841(n)*A322086(n) mod 13^n.

a(n) == L(13^n,7) (mod 13^n) == ((7 + sqrt(53))/2)^(13^n) + ((7 - sqrt(53))/2)^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial, the n-th row polynomial of A114525. - Peter Bala, Dec 05 2022

Example

7^2 = 49 = 4*13 - 3.
124^2 = 15376 = 91*13^2 - 3 = 7*13^3 - 3.
13306^2 = 177049636 = 6199*13^4 - 3.

Prog

(PARI) a(n) = truncate(-sqrt(-3+O(13^n)))

Crossrefs

Cf. A114525, A286840, A286841, A322085, A322086, A322089, A322092.

Sequence in context: A012086 A074487 A192566 * A360337 A304420 A217910

Adjacent sequences: A322087 A322088 A322089 * A322091 A322092 A322093

Keyword

nonn,easy

Author

Jianing Song, Nov 26 2018

Status

approved