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%I #14 Jul 12 2022 17:51:12
%S 1,7,4,8,8,2,8,6,3,7,3,6,6,1,7,8,5,8,9,7,2,8,7,7,5,3,8,3,9,8,9,8,7,2,
%T 7,1,7,1,1,6,3,2,9,2,2,2,7,7,3,7,3,0,0,3,1,8,6,7,8,4,5,6,2,5,2,2,3,0,
%U 3,8,5,9,7,9,0,3,6,3,3,8,0,8,0,0,2,5,0,1,1,2,2,6,9,1,2,2,1,1,9,1,8,8,5,7,7
%N a(n)...a(1) = digital representation of n-digit number m, the cube of which, m^3, ends with n 1's.
%C For any n there is only one solution. Case a(n)=0 means that cube of (n-1)-digit number ends with n (not (n-1)) 1's. Case a(n+1)=a(n)=0 means that cube of (n-1)-digit number ends with (n+1) (not (n-1)) 1's, etc.
%C 10-adic integer x such that x^3 == (10^n-1)/9 mod 10^n. - _Aswini Vaidyanathan_, May 07 2013
%C 10-adic digits of the cubic root of -1/9. - _Max Alekseyev_, Jul 12 2022
%H Robert Israel, <a href="/A153042/b153042.txt">Table of n, a(n) for n = 1..10000</a>
%e 1^3= 1; 71^3 = 357911; 471^3 = 104487111; 8471^3 = 607860671111.
%p N:= 200:
%p op([1,3], padic:-rootp(9*x^3+1,10,N+2))[1..N+1]; # _Robert Israel_, Mar 25 2018
%o (PARI) n=0; for(i=1, 100, m=(10^i-1)/9; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break))) \\ _Aswini Vaidyanathan_, May 07 2013
%o (PARI) digits(sqrtn(-1/9 + O(10^100),3)) \\ _Max Alekseyev_, Jul 12 2022
%K base,nonn
%O 1,2
%A _Zak Seidov_, Dec 17 2008, corrected Dec 20 2008
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