A324085 - OEIS

%I #9 Sep 07 2019 19:09:26

%S 2,2,7,1,12,6,12,4,8,6,1,10,4,6,7,8,10,1,12,9,9,7,0,12,3,6,4,5,11,12,

%T 3,11,9,5,8,4,4,2,7,4,11,8,4,10,1,0,2,1,4,3,11,7,3,6,3,2,6,7,3,6,1,0,

%U 3,0,11,8,11,6,11,0,3,5,4,7,9,10,12,6,11,5,1

%N Digits of one of the four 3-adic integers 3^(1/4) that is congruent to 2 mod 13.

%C One of the two square roots of A322087, where an A-number represents a 13-adic number. The other square root is A324153.

%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 A324086*A286838 = A324087*A286839.

%F a(n) = (A324077(n+1) - A324077(n))/13^n.

%F For n > 0, a(n) = 12 - A324153(n).

%e 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 = (722)_13, so the first three terms are 2, 7 and 7.

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

%Y Cf. A286838, A286839, A322087, A324077, A324082, A324083, A324084, A324086, A324087, A324153.

%K nonn,base

%O 0,1

%A _Jianing Song_, Sep 01 2019