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A153042
a(n)...a(1) = digital representation of n-digit number m, the cube of which, m^3, ends with n 1's.
9
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
1,2
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
LINKS
EXAMPLE
1^3= 1; 71^3 = 357911; 471^3 = 104487111; 8471^3 = 607860671111.
MAPLE
N:= 200:
op([1, 3], padic:-rootp(9*x^3+1, 10, N+2))[1..N+1]; # Robert Israel, Mar 25 2018
PROG
(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
CROSSREFS
Sequence in context: A330596 A296427 A092034 * A330595 A119824 A271173
KEYWORD
base,nonn
AUTHOR
Zak Seidov, Dec 17 2008, corrected Dec 20 2008
STATUS
approved