A153042 a(n)...a(1) = digital representation of n-digit number m, the cube of which, m^3, ends with n 1'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, 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, 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

Offset: 1

Author

Zak Seidov, Dec 17 2008, corrected Dec 20 2008

Keywords

• base
• nonn

Comments

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.

10-adic integer x such that x^3 == (10^n-1)/9 mod 10^n. - Aswini Vaidyanathan, May 07 2013

10-adic digits of the cubic root of -1/9. - Max Alekseyev, Jul 12 2022

Examples

			1^3= 1; 71^3 = 357911; 471^3 = 104487111; 8471^3 = 607860671111.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Programs

Maple

N:= 200:
op([1,3], padic:-rootp(9*x^3+1,10,N+2))[1..N+1]; # Robert Israel, Mar 25 2018

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

PARI

digits(sqrtn(-1/9 + O(10^100),3)) \\ Max Alekseyev, Jul 12 2022