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%I #8 Sep 07 2019 19:09:38
%S 10,7,9,6,7,0,2,10,0,0,4,0,1,5,12,10,7,1,1,9,7,1,7,8,0,0,9,10,5,5,0,1,
%T 4,7,0,9,7,4,6,0,3,8,12,7,7,0,11,3,11,3,1,5,8,12,9,3,12,0,6,6,11,4,8,
%U 3,7,6,3,7,5,2,11,9,9,4,7,1,4,10,12,11,0
%N Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 10 mod 13.
%C One of the two square roots of A322088, where an A-number represents a 13-adic number. The other square root is A324086.
%C For k not divisible by 13, k is a fourth power in 13-adic field if and only if k == 1, 3, 9 (mod 13). If k is a fourth power in 13-adic field, then k has exactly 4 fourth-power roots.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F Equals A324085*A286838 = A324153*A286839.
%F a(n) = (A324083(n+1) - A324083(n))/13^n.
%F For n > 0, a(n) = 12 - A324086(n).
%e The unique number k in [1, 13^3] and congruent to 10 modulo 13 such that k^4 - 3 is divisible by 13^3 is k = 1622 = (97A)_13, so the first three terms are 10, 7 and 9.
%o (PARI) a(n) = lift(-sqrtn(3+O(13^(n+1)), 4))\13^n
%Y Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324085, A324086, A324153.
%K nonn,base
%O 0,1
%A _Jianing Song_, Sep 01 2019
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