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%I #44 Aug 30 2019 02:43:43
%S 0,1,1,1,9,25,25,25,25,281,281,281,281,4377,4377,20761,53529,53529,
%T 184601,446745,971033,2019609,4116761,8311065,8311065,25088281,
%U 58642713,125751577,259969305,259969305,259969305,259969305,259969305,4554936601,13144871193
%N The successive approximations up to 2^n for 2-adic integer 9^(1/3).
%C a(n) is the unique solution to x^3 == 9 (mod 2^n) in the range [0, 2^n - 1].
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>
%F For n > 0, a(n) = a(n-1) if a(n-1)^3 - 9 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
%e 9^3 = 729 = 45*2^4 + 9;
%e 25^3 = 15625 = 488*2^5 + 9 = 244*2^6 + 9 = 122*2^7 + 9 = 61*2^8 + 9;
%e 281^3 = 22188041 = 43336*2^9 + 9 = 21668*2^10 + 9 = 10834*2^11 + 9 = 5417*2^12 + 9.
%o (PARI) a(n) = lift(sqrtn(9+O(2^n), 3))
%Y For the digits of 9^(1/3), see A323096.
%Y Approximations of p-adic cubic roots:
%Y A322701 (2-adic, 3^(1/3));
%Y A322926 (2-adic, 5^(1/3));
%Y A322934 (2-adic, 7^(1/3));
%Y this sequence (2-adic, 9^(1/3));
%Y A290567 (5-adic, 2^(1/3));
%Y A290568 (5-adic, 3^(1/3));
%Y A309444 (5-adic, 4^(1/3));
%Y A319097, A319098, A319199 (7-adic, 6^(1/3));
%Y A320914, A320915, A321105 (13-adic, 5^(1/3)).
%K nonn
%O 0,5
%A _Jianing Song_, Aug 30 2019
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