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10-adic integer x such that x^9 = 1/3.
1

%I #12 Aug 16 2019 15:26:12

%S 7,4,1,1,5,6,4,8,1,0,5,6,9,7,1,2,6,3,4,0,0,8,5,0,1,9,6,3,9,3,2,5,1,3,

%T 4,7,5,4,3,1,9,0,7,8,3,7,0,7,9,9,3,7,5,9,5,3,8,9,0,8,8,2,9,4,4,1,3,3,

%U 8,0,5,6,0,4,3,2,6,1,2,6,2,4,4,6,5,7,4,0,6,1,7,2,2,6,0,1,2,4,9,5

%N 10-adic integer x such that x^9 = 1/3.

%C This is the 10's complement of A225454.

%H Seiichi Manyama, <a href="/A225463/b225463.txt">Table of n, a(n) for n = 0..10000</a>

%e 7^9 == -3 (mod 10).

%e 47^9 == -33 (mod 10^2).

%e 147^9 == -333 (mod 10^3).

%e 1147^9 == -3333 (mod 10^4).

%e 51147^9 == -33333 (mod 10^5).

%e 651147^9 == -333333 (mod 10^6).

%o (PARI) n=0;for(i=1,100,m=(2*(10^i-1)/3)+1;for(x=0,9,if(((n+(x*10^(i-1)))^9)%(10^i)==m,n=n+(x*10^(i-1));print1(x", ");break)))

%o (PARI) N=100; Vecrev(digits(lift(chinese(Mod((1/3+O(2^N))^(1/9), 2^N), Mod((1/3+O(5^N))^(1/9), 5^N)))), N) \\ _Seiichi Manyama_, Aug 07 2019

%Y Cf. A225454.

%K nonn,base

%O 0,1

%A _Aswini Vaidyanathan_, May 11 2013