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%I #14 Aug 28 2019 14:58:38
%S 8,0,1,5,7,0,5,12,8,10,11,6,9,3,4,5,8,1,5,3,0,7,1,2,7,8,8,3,4,1,0,11,
%T 4,0,0,5,4,7,2,9,4,3,4,11,11,6,8,12,11,5,2,1,7,12,7,7,11,11,0,6,5,9,6,
%U 12,5,3,11,5,12,4,9,5,1,9,9,3,8,0,7,0,3
%N Digits of one of the three 13-adic integers 5^(1/3) that is related to A320915.
%C For k not divisible by 5, 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 Seiichi Manyama, <a href="/A321107/b321107.txt">Table of n, a(n) for n = 0..10000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F a(n) = (A320915(n+1) - A320915(n))/13^n.
%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 = (108)_13, so the first three terms are 8, 0 and 1.
%o (PARI) a(n) = lift(sqrtn(5+O(13^(n+1)), 3))\13^n
%Y Cf. A320914, A320915, A321105, A321106, A321108.
%Y For 5-adic cubic roots, see A290566, A290563, A309443.
%K nonn,base
%O 0,1
%A _Jianing Song_, Aug 27 2019