A322089 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-3). Here the 6 (mod 13) case (except for n = 0).

0, 6, 45, 2073, 15255, 300865, 2899916, 22207152, 273201220, 7614777709, 92450772693, 1333177199334, 4917497987408, 191302178967256, 1705677711928521, 48954194340319989, 202511873382592260, 3529594919298491465, 38131258596823843197, 38131258596823843197, 8809653000849500507259

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 6 modulo 13.

A322090 is the approximation (congruent to 7 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 - A322090(n).

a(n) = Sum_{i=0..n-1} A322091(i)*13^i.

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

a(n) == L(13^n,6) (mod 13^n) == (3 + sqrt(10))^(13^n) + (3 - sqrt(10))^(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

6^2 = 36 = 3*13 - 3.
45^2 = 2025 = 12*13^2 - 3.
2073^2 = 4297329 = 1956*13^3 - 3.

Prog

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

Crossrefs

Cf. A114525, A286840, A286841, A322085, A322086, A322090, A322091.

Sequence in context: A160492 A318017 A356487 * A273091 A086721 A145002

Adjacent sequences: A322086 A322087 A322088 * A322090 A322091 A322092

Keyword

nonn,easy

Author

Jianing Song, Nov 26 2018

Status

approved