A322999 The successive approximations up to 2^n for 2-adic integer 9^(1/3).

0, 1, 1, 1, 9, 25, 25, 25, 25, 281, 281, 281, 281, 4377, 4377, 20761, 53529, 53529, 184601, 446745, 971033, 2019609, 4116761, 8311065, 8311065, 25088281, 58642713, 125751577, 259969305, 259969305, 259969305, 259969305, 259969305, 4554936601, 13144871193

Offset: 0

Author

Jianing Song, Aug 30 2019

Keywords

• nonn

Comments

a(n) is the unique solution to x^3 == 9 (mod 2^n) in the range [0, 2^n - 1].

Examples

			9^3 = 729 = 45*2^4 + 9;
25^3 = 15625 = 488*2^5 + 9 = 244*2^6 + 9 = 122*2^7 + 9 = 61*2^8 + 9;
281^3 = 22188041 = 43336*2^9 + 9 = 21668*2^10 + 9 = 10834*2^11 + 9 = 5417*2^12 + 9.
		

Links

• Wikipedia, p-adic number

Crossrefs

For the digits of 9^(1/3), see A323096.

Approximations of p-adic cubic roots:

A322701 (2-adic, 3^(1/3));

A322926 (2-adic, 5^(1/3));

A322934 (2-adic, 7^(1/3));

this sequence (2-adic, 9^(1/3));

A290567 (5-adic, 2^(1/3));

A290568 (5-adic, 3^(1/3));

A309444 (5-adic, 4^(1/3));

A319097, A319098, A319199 (7-adic, 6^(1/3));

A320914, A320915, A321105 (13-adic, 5^(1/3)).

Programs

PARI

a(n) = lift(sqrtn(9+O(2^n), 3))

Formula

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).