A322085 One of the two successive approximations up to 13^n for 13-adic integer sqrt(3). Here the 4 (mod 13) case (except for n = 0).

0, 4, 108, 1122, 18698, 361430, 1104016, 5930825, 570667478, 7912243967, 113957237697, 251815729546, 11004778093768, 104197118583692, 3132948184506222, 26757206498701956, 589802029653700283, 7909384730668678534, 85763128005100719931, 648040162764887685576

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 4 modulo 13.

A322086 is the approximation (congruent to 9 mod 13) of another square root of 3 over the 13-adic field.

Examples

			4^2 = 16 = 1*13 + 3.
108^2 = 11664 = 69*13^2 + 3.
1122^2 = 1258884 = 573*13^3 + 3.
		

Links

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

Crossrefs

Cf. A286840, A286841, A322086, A322087, A322089, A322090.

Programs

Maple

S:= map(t -> op([1,3],t),[padic:-evalp(RootOf(x^2-3,x),13,30)]):
S4:= op(select(t -> t[1]=4, S)):
seq(add(S4[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 - A322086(n).

a(n) = Sum_{i=0..n-1} A322087(i)*13^i.

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