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%I #44 Aug 29 2019 10:23:37
%S 0,7,7,1021,20794,77916,4533432,57628331,810610535,8967917745,
%T 40781415864,592215383260,22098140111704,208482821091552,
%U 3842984100198588,23529866028695033,586574689183693360,5244490953465952247,74447818308516655711,524269446116346228227,9295791188369022892289
%N One of the three successive approximations up to 13^n for 13-adic integer 5^(1/3). This is the 7 (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 7 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 7 modulo 13 such that k^3 - 5 is divisible by 13^2 is k = 7, so a(2) = 7.
%e The unique number k in [1, 13^3] and congruent to 7 modulo 13 such that k^3 - 5 is divisible by 13^3 is k = 1021, so a(3) = 1021.
%o (PARI) a(n) = lift(sqrtn(5+O(13^n), 3) * (-1+sqrt(-3+O(13^n)))/2)
%Y Cf. A320915, 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
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