A324083 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 10 (mod 13) case (except for n = 0).

0, 10, 101, 1622, 14804, 214731, 214731, 9868349, 637353519, 637353519, 637353519, 552071320915, 552071320915, 23850156443396, 1538225689404661, 48786742317796129, 560645672458703699, 5218561936740962586, 13868977856122300519, 126324384808079693648

Offset: 0

Author

Jianing Song, Sep 01 2019

Keywords

• nonn

Comments

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 10 mod 13 such that k^4 - 3 is divisible by 13^n.

For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.

Examples

			The unique number k in [1, 13^2] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 101, so a(2) = 101.
The unique number k in [1, 13^3] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1622, so a(3) = 1622.
		

Links

• Wikipedia, p-adic number

Crossrefs

Cf. A286840, A286841, A322085, A324077, A324082, A324084, A324085, A324086, A324087, A324153.

Programs

PARI

a(n) = lift(-sqrtn(3+O(13^n), 4))

Formula

a(n) = A324077(n)*A286840(n) mod 13^n = A324084(n)*A286841(n) mod 13^n.

For n > 0, a(n) = 13^n - A324082(n).

a(n)^2 == A322086(n) (mod 13^n).