

A324077


One of the four successive approximations up to 13^n for 13adic integer 3^(1/4).This is the 2 (mod 13) case (except for n = 0).


11



0, 2, 28, 1211, 3408, 346140, 2573898, 60495606, 311489674, 6837335442, 70464331680, 208322823529, 18129926763899, 111322267253823, 1928572906807341, 29490207606702364, 438977351719428420, 7093143443551226830, 15743559362932564763, 1365208442786421282311
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OFFSET

0,2


COMMENTS

For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 2 mod 13 such that k^4  3 is divisible by 13^n.
For k not divisible by 13, k is a fourth power in 13adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13adic field, then k has exactly 4 fourthpower roots.


LINKS

Table of n, a(n) for n=0..19.
Wikipedia, padic number


FORMULA

a(n) = A324082(n)*A286840(n) mod 13^n = A324083(n)*A286841(n) mod 13^n.
For n > 0, a(n) = 13^n  A324084(n).
a(n)^2 == A322085(n) (mod 13^n).


EXAMPLE

The unique number k in [1, 13^2] and congruent to 2 modulo 13 such that k^4  3 is divisible by 13^2 is k = 28, so a(2) = 28.
The unique number k in [1, 13^3] and congruent to 2 modulo 13 such that k^4  3 is divisible by 13^3 is k = 1211, so a(3) = 1211.


PROG

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


CROSSREFS

Cf. A286840, A286841, A322085, A324082, A324083, A324084, A324085, A324086, A324087, A324153.
Sequence in context: A300459 A009674 A143598 * A071220 A063794 A238817
Adjacent sequences: A324074 A324075 A324076 * A324078 A324079 A324080


KEYWORD

nonn


AUTHOR

Jianing Song, Sep 01 2019


STATUS

approved



