A325485 One of the four successive approximations up to 5^n for the 5-adic integer 6^(1/4). This is the 2 (mod 5) case (except for n = 0).

0, 2, 22, 22, 397, 397, 6647, 6647, 319147, 319147, 6178522, 6178522, 103834772, 592116022, 3033522272, 9137037897, 70172194147, 222760084772, 3274517897272, 3274517897272, 60494976881647, 441964703444147, 1395639019850397, 3779824810866022, 51463540631178522

Offset: 0

Author

Jianing Song, Sep 07 2019

Keywords

• nonn

Comments

For n > 0, a(n) is the unique number k in [1, 5^n] and congruent to 2 mod 5 such that k^4 - 6 is divisible by 5^n.

For k not divisible by 5, k is a fourth power in 5-adic field if and only if k == 1 (mod 5). If k is a fourth power in 5-adic field, then k has exactly 4 fourth-power roots.

Examples

			The unique number k in [1, 5^2] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^2 is k = 22, so a(2) = 22.
The unique number k in [1, 5^3] and congruent to 2 modulo 5 such that k^4 - 6 is divisible by 5^3 is also k = 22, so a(3) is also 22.
		

Links

• Wikipedia, p-adic number

Crossrefs

Cf. A048898, A048899, A324024, A325489, A325490, A325491, A325492.

Approximations of p-adic fourth-power roots:

A325484, this sequence, A325486, A325487 (5-adic, 6^(1/4));

A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

Programs

PARI

a(n) = lift(sqrtn(6+O(5^n), 4) * sqrt(-1+O(5^n)))

Formula

a(n) = A325484(n)*A048898(n) mod 13^n = A325485(n)*A048899(n) mod 13^n.

For n > 0, a(n) = 5^n - A325486(n).

a(n)^2 == A324024(n) (mod 5^n).