login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Successive approximations up to 2^n for the 2-adic integer 17^(1/4). This is the 1 (mod 4) case.
3

%I #10 Feb 19 2021 12:39:01

%S 1,5,13,13,45,45,173,429,429,1453,3501,7597,7597,23981,23981,23981,

%T 155053,417197,941485,1990061,1990061,1990061,10378669,10378669,

%U 10378669,10378669,144596397,413031853,413031853,413031853,2560515501,6855482797,15445417389,15445417389

%N Successive approximations up to 2^n for the 2-adic integer 17^(1/4). This is the 1 (mod 4) case.

%C a(n) is the unique number k in [1, 2^n] and congruent to 1 mod 4 such that k^4 - 17 is divisible by 2^(n+2).

%C For odd k, k has a fourth root in the ring of 2-adic integers if and only if k == 1 (mod 16), in which case k has exactly two fourth roots.

%H Jianing Song, <a href="/A341751/b341751.txt">Table of n, a(n) for n = 2..1000</a>

%F a(2) = 1; for n >= 3, a(n) = a(n-1) if a(n-1)^4 - 17 is divisible by 2^(n+2), otherwise a(n-1) + 2^(n-1).

%F a(n) = 2^n - A341752(n).

%F a(n) = Sum_{i=0..n-1} A341753(i)*2^i.

%F a(n)^2 == A341538(n) (mod 2^n).

%e The unique number k in [1, 4] and congruent to 1 modulo 4 such that k^4 - 17 is divisible by 16 is 1, so a(2) = 1.

%e a(2)^4 - 17 = -16 which is not divisible by 32, so a(3) = a(2) + 2^2 = 5.

%e a(3)^4 - 17 = 608 which is not divisible by 64, so a(4) = a(3) + 2^3 = 13.

%e a(4)^4 - 17 = 28544 which is divisible by 128, so a(5) = a(4) = 13.

%e a(5)^4 - 17 = 28544 which is not ndivisible by 256, so a(6) = a(5) + 2^5 = 45.

%e ...

%o (PARI) a(n) = truncate(sqrtn(17+O(2^(n+2)), 4))

%Y Cf. A341753 (digits of the associated 2-adic fourth root of 17), A341538.

%Y Approximations of p-adic fourth-power roots:

%Y this sequence, A341752 (2-adic, 17^(1/4));

%Y A325484, A325485, A325486, A325487 (5-adic, 6^(1/4));

%Y A324077, A324082, A324083, A324084 (13-adic, 3^(1/4)).

%K nonn

%O 2,2

%A _Jianing Song_, Feb 18 2021