A327303 - OEIS

%I #19 Jan 17 2023 06:26:21

%S 0,4,4,79,79,79,3204,18829,331329,1112579,1112579,20643829,118300079,

%T 850721954,3292128204,27706190704,149776503204,302364393829,

%U 1065303846954,8694698378204,46841671034454,332943965956329,332943965956329,5101315547987579,28943173458143829

%N One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-9). This is the 4 (mod 5) case (except for n = 0).

%C a(n) is the unique number k in [1, 5^n] and congruent to 4 mod 5 such that k^2 + 9 is divisible by 5^n.

%H Robert Israel, <a href="/A327303/b327303.txt">Table of n, a(n) for n = 0..1424</a>

%H G. P. Michon, <a href="http://www.numericana.com/answer/p-adic.htm#integers">Introduction to p-adic integers</a>, Numericana.

%F a(1) = 4; for n >= 2, a(n) is the unique number k in {a(n-1) + m*5^(n-1) : m = 0, 1, 2, 3, 4} such that k^2 + 9 is divisible by 5^n.

%F For n > 0, a(n) = 5^n - A327302(n).

%e The unique number k in {4, 9, 14, 19, 24} such that k^2 + 9 is divisible by 25 is k = 4, so a(2) = 4.

%e The unique number k in {4, 29, 54, 79, 104} such that k^2 + 9 is divisible by 125 is k = 79, so a(3) = 46.

%e The unique number k in {79, 204, 329, 454, 579} such that k^2 + 9 is divisible by 625 is k = 79, so a(4) = 79.

%p R:= [padic:-rootp(x^2+9,5,101)]:

%p R:= op(select(t -> padic:-ratvaluep(t,1)=4, R)):

%p seq(padic:-ratvaluep(R,n),n=0..100); # _Robert Israel_, Jan 16 2023

%o (PARI) a(n) = truncate(-sqrt(-9+O(5^n)))

%Y For the digits of sqrt(-9) see A327304 and A327305.

%Y Approximations of 5-adic square roots:

%Y A327302, this sequence (sqrt(-9));

%Y A324027, A324028 (sqrt(-6));

%Y A268922, A269590 (sqrt(-4));

%Y A048898, A048899 (sqrt(-1));

%Y A324023, A324024 (sqrt(6)).

%K nonn

%O 0,2

%A _Jianing Song_, Sep 16 2019