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A324077 One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 2 (mod 13) case (except for n = 0). 13

%I #31 Sep 07 2019 19:08:41

%S 0,2,28,1211,3408,346140,2573898,60495606,311489674,6837335442,

%T 70464331680,208322823529,18129926763899,111322267253823,

%U 1928572906807341,29490207606702364,438977351719428420,7093143443551226830,15743559362932564763,1365208442786421282311

%N One of the four successive approximations up to 13^n for 13-adic integer 3^(1/4).This is the 2 (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 2 mod 13 such that k^4 - 3 is divisible by 13^n.

%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 a(n) = A324082(n)*A286840(n) mod 13^n = A324083(n)*A286841(n) mod 13^n.

%F For n > 0, a(n) = 13^n - A324084(n).

%F a(n)^2 == A322085(n) (mod 13^n).

%e The unique number k in [1, 13^2] and congruent to 2 modulo 13 such that k^4 - 3 is divisible by 13^2 is k = 28, so a(2) = 28.

%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, so a(3) = 1211.

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

%Y Cf. A286840, A286841, A322085, A324082, A324083, A324084, A324085, A324086, A324087, A324153.

%K nonn

%O 0,2

%A _Jianing Song_, Sep 01 2019

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Last modified April 16 17:08 EDT 2024. Contains 371749 sequences. (Running on oeis4.)