OFFSET
0,1
COMMENTS
This is the 10's complement of A153042.
Equivalently, the 10-adic cube root of 1/9, i.e., x such that 9*x^3 = 1 (mod 10^n) for all n. - M. F. Hasler, Jan 02 2019
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..10000
FORMULA
p = ...711529, q = A153042 = ...288471, p + q = 0. - Seiichi Manyama, Aug 04 2019
Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 9, b(n) = b(n-1) + 7 * (9 * b(n-1)^3 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019
EXAMPLE
9^3 == -1 (mod 10).
29^3 == -11 (mod 10^2).
529^3 == -111 (mod 10^3).
1529^3 == -1111 (mod 10^4).
11529^3 == -11111 (mod 10^5).
711529^3 == -111111 (mod 10^6).
MAPLE
op([1, 3], padic:-rootp(9*x^3 -1, 10, 101)); # Robert Israel, Aug 04 2019
PROG
(PARI) n=0; for(i=1, 100, m=(8*(10^i-1)/9)+1; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))
(PARI) upto(N=100, m=1/3)=Vecrev(digits(lift(chinese(Mod((1/9+O(5^N))^m, 5^N), Mod((1/9+O(2^N))^m, 2^N)))), N) \\ Following Andrew Howroyd's code for A319740. - M. F. Hasler, Jan 02 2019
(PARI) Vecrev(digits(truncate(-(-1/9+O(10^100))^(1/3)))) \\ Seiichi Manyama, Aug 04 2019
(Ruby)
def A225412(n)
ary = [9]
a = 9
n.times{|i|
b = (a + 7 * (9 * a ** 3 - 1)) % (10 ** (i + 2))
ary << (b - a) / (10 ** (i + 1))
a = b
}
ary
end
p A225412(100) # Seiichi Manyama, Aug 13 2019
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Aswini Vaidyanathan, May 07 2013
EXTENSIONS
Name edited by Seiichi Manyama, Aug 04 2019
STATUS
approved