A325892 The successive approximations up to 2^n for the 2-adic integer 3^(1/5).
0, 1, 3, 3, 3, 19, 19, 83, 211, 211, 211, 211, 2259, 6355, 14547, 30931, 63699, 129235, 129235, 129235, 129235, 129235, 129235, 4323539, 4323539, 4323539, 4323539, 4323539, 4323539, 4323539, 4323539, 4323539, 2151807187, 6446774483, 6446774483, 6446774483
Offset: 0
Keywords
Examples
For n = 2, the unique solution to x^5 == 3 (mod 4) in the range [0, 3] is x = 3, so a(2) = 3. a(2)^5 - 3 = 240 which is divisible by 8, so a(3) = a(2) = 3; a(3)^5 - 3 = 240 which is divisible by 16, so a(4) = a(3) = 3; a(4)^5 - 3 = 240 which is not divisible by 32, so a(5) = a(4) + 16 = 19; a(5)^5 - 3 = 2476096 which is divisible by 64, so a(6) = a(5) = 19.
Links
- Wikipedia, p-adic number
Crossrefs
For the digits of 3^(1/5), see A325896.
Approximations of p-adic fifth-power roots:
this sequence (2-adic, 3^(1/5));
A325893 (2-adic, 5^(1/5));
A325894 (2-adic, 7^(1/5));
A325895 (2-adic, 9^(1/5));
A322157 (5-adic, 7^(1/5));
A309450 (7-adic, 2^(1/5));
A309451 (7-adic, 3^(1/5));
A309452 (7-adic, 4^(1/5));
A309453 (7-adic, 5^(1/5));
A309454 (7-adic, 6^(1/5)).
Programs
-
PARI
a(n) = lift(sqrtn(3+O(2^n), 5))
Formula
For n > 0, a(n) = a(n-1) if a(n-1)^5 - 3 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
Comments