A322087 Digits of one of the two 13-adic integers sqrt(3).

4, 8, 6, 8, 12, 2, 1, 9, 9, 10, 1, 6, 4, 10, 6, 11, 11, 9, 5, 5, 0, 5, 2, 5, 8, 0, 8, 7, 3, 5, 3, 12, 0, 3, 10, 3, 5, 8, 1, 12, 11, 8, 7, 0, 3, 1, 4, 9, 9, 9, 1, 10, 6, 12, 2, 7, 3, 5, 1, 6, 12, 1, 1, 12, 10, 5, 6, 11, 7, 8, 12, 10, 1, 3, 5, 5, 5, 7, 11, 1, 5

Offset

0,1

Comments

This square root of 3 in the 13-adic field ends with digit 4. The other, A322088, ends with digit 9.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Peter Bala, Using Chebyshev polynomials to find the p-adic square roots of 2 and 3, Dec 2022.

Wikipedia, p-adic number

Formula

a(n) = (A322085(n+1) - A322085(n))/13^n.

For n > 0, a(n) = 12 - A322088(n).

Equals A286838*A322091 = A286839*A322092.

This 13-adic integer is the 13-adic limit as n -> oo of the integer sequence {2*T(13^n,2)}, where T(n,x) denotes the n-th Chebyshev polynomial. - Peter Bala, Dec 04 2022

Example

...BC1853A30C35378085250559BB6A461A9912C8684.

Prog

(PARI) a(n) = truncate(sqrt(3+O(13^(n+1))))\13^n

Crossrefs

Cf. A322085.

Digits of p-adic integers:

A321074, A321075 (11-adic, sqrt(3));

this sequence, A322088 (13-adic, sqrt(3));

A286838, A286839 (13-adic, sqrt(-1));

A322091, A322092 (13-adic, sqrt(-3)).

Sequence in context: A279160 A197290 A200368 * A142350 A011515 A005531

Adjacent sequences: A322084 A322085 A322086 * A322088 A322089 A322090

Keyword

nonn,base

Author

Jianing Song, Nov 26 2018

Status

approved