A322086 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 9 (mod 13) case (except for n = 0).

0, 9, 61, 1075, 9863, 9863, 3722793, 56817692, 245063243, 2692255406, 23901254152, 1540344664491, 12293307028713, 198677988008561, 804428201193067, 24428686515388801, 75614579529479558, 741031188712659399, 26692278946856673198, 813880127610558425101, 11047322160238681199840

Offset: 0

Author

Jianing Song, Nov 26 2018

Keywords

• nonn

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 9 modulo 13.

A322085 is the approximation (congruent to 4 mod 13) of another square root of 3 over the 13-adic field.

Examples

			9^2 = 81 = 6*13 + 3.
61^2 = 3721 = 22*13^2 + 3.
1075^2 = 1155625 = 526*13^3 + 3.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..896
• Wikipedia, p-adic number

Crossrefs

Cf. A286840, A286841, A322085, A322088, A322089, A322090.

Programs

Maple

S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
S9:= op(select(t -> t[1]=9, S)):
seq(add(S9[i]*13^(i-1),i=1..n-1),n=1..31); # Robert Israel, Jun 13 2019

PARI

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

Formula

For n > 0, a(n) = 13^n - A322085(n).

a(n) = Sum_{i=0..n-1} A322088(i)*13^i.

a(n) = A286840(n)*A322090(n) mod 13^n = A286841(n)*A322089(n) mod 13^n.