A321073 One of the two successive approximations up to 11^n for 11-adic integer sqrt(3). Here the 6 (mod 11) case (except for n = 0).

0, 6, 94, 578, 3240, 135009, 296060, 2067621, 118990647, 1619502814, 3977450505, 211476847313, 782100188535, 22751098825582, 229887371689168, 609637205272409, 38204870730013268, 84154600593585429, 3622283800088641826, 42541704994534262193, 654132609478679725103

Offset

0,2

Comments

For n > 0, a(n) is the unique solution to x^2 == 3 (mod 11^n) in the range [0, 11^n - 1] and congruent to 6 modulo 11.

A321072 is the approximation (congruent to 5 mod 11) of another square root of 3 over the 11-adic field.

Links

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

Wikipedia, p-adic number

Formula

For n > 0, a(n) = 11^n - A321072(n).

a(n) = Sum_{i=0..n-1} A321075(i)*11^i.

a(n) == (3 + sqrt(8))^(11^n) + (3 - sqrt(8))^(11^n) (mod 11^n). - Peter Bala, Dec 04 2022

Example

6^2 = 36 = 3 + 3*11.
94^2 = 8836 = 3 + 73*11^2.
578^2 = 334084 = 3 + 251*11^3.

Prog

(PARI) a(n) = truncate(-sqrt(3+O(11^n)))

Crossrefs

Cf. A321072, A321075.

Sequence in context: A218682 A078103 A221525 * A198257 A296820 A184983

Adjacent sequences: A321070 A321071 A321072 * A321074 A321075 A321076

Keyword

nonn,easy

Author

Jianing Song, Oct 27 2018

Status

approved