%I #65 Aug 29 2019 10:23:33

%S 0,8,8,177,11162,211089,211089,24345134,777327338,7303173106,

%T 113348166836,1629791577175,12382753941397,222065520043726,

%U 1130690839820485,16880196382617641,272809661453071426,5596142534918510154,14246558454299848087,576523593214086813732,4962284464340425145763

%N One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 8 (mod 13) case (except for n = 0).

%C For n > 0, a(n) is the unique number k in [1, 13^n] and congruent to 8 mod 13 such that k^3 - 5 is divisible by 13^n.

%C For k not divisible by 13, k is a cube in 13-adic field if and only if k == 1, 5, 8, 12 (mod 13). If k is a cube in 13-adic field, then k has exactly three cubic roots.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

%e The unique number k in [1, 13^2] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 8, so a(2) = 8.

%e The unique number k in [1, 13^3] and congruent to 8 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 177, so a(3) = 177.

%o (PARI) a(n) = lift(sqrtn(5+O(13^n), 3))

%Y Cf. A320914, A321105, A321106, A321107, A321108.

%Y For 5-adic cubic roots, see A290567, A290568, A309444.

%K nonn

%O 0,2

%A _Jianing Song_, Aug 27 2019