A325895 The successive approximations up to 2^n for the 2-adic integer 9^(1/5).
0, 1, 1, 1, 9, 9, 41, 105, 233, 489, 489, 1513, 3561, 7657, 15849, 32233, 32233, 97769, 228841, 490985, 1015273, 2063849, 4161001, 8355305, 16743913, 16743913, 16743913, 83852777, 218070505, 218070505, 218070505, 1291812329, 1291812329, 5586779625, 5586779625, 5586779625
Offset: 0
Keywords
Examples
For n = 2, the unique solution to x^5 == 9 (mod 4) in the range [0, 3] is x = 1, so a(2) = 1. a(2)^5 - 9 = -8 which is divisible by 8, so a(3) = a(2) = 1; a(3)^5 - 9 = -8 which is not divisible by 16, so a(4) = a(3) + 8 = 9; a(4)^5 - 9 = 59040 which is divisible by 32, so a(5) = a(4) = 9; a(5)^5 - 9 = 59040 which is not divisible by 64, so a(6) = a(5) + 32 = 41.
Links
- Wikipedia, p-adic number
Crossrefs
For the digits of 9^(1/5), see A325899.
Approximations of p-adic fifth-power roots:
A325892 (2-adic, 3^(1/5));
A325893 (2-adic, 5^(1/5));
A325894 (2-adic, 7^(1/5));
this sequence (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
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PARI
a(n) = lift(sqrtn(9+O(2^n), 5))
Formula
For n > 0, a(n) = a(n-1) if a(n-1)^5 - 9 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
Comments