A324084 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 11 (mod 13) case (except for n = 0).

0, 11, 141, 986, 25153, 25153, 2252911, 2252911, 504241047, 3767163931, 67394160169, 1583837570508, 5168158358582, 191552839338430, 2008803478891948, 21695685407388393, 226439257463751421, 1557272475830111103, 96711847589024828366, 96711847589024828366

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 11 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 11 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 141, so a(2) = 141.
The unique number k in [1, 13^3] and congruent to 11 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 986, so a(3) = 986.
		

Links

• Wikipedia, p-adic number

Crossrefs

Cf. A286840, A286841, A322085, A324077, A324082, A324083, A324085, A324086, A324087, A324153.

Programs

PARI

a(n) = lift(-sqrtn(3+O(13^n), 4) * sqrt(-1+O(13^n)))

Formula

a(n) = A324082(n)*A286841(n) mod 13^n = A324083(n)*A286840(n) mod 13^n.

For n > 0, a(n) = 13^n - A324077(n).

a(n)^2 == A322085(n) (mod 13^n).