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A225412
Digits of the 10-adic integer (1/9)^(1/3).
10
9, 2, 5, 1, 1, 7, 1, 3, 6, 2, 6, 3, 3, 8, 2, 1, 4, 1, 0, 2, 7, 1, 2, 2, 4, 6, 1, 6, 0, 1, 0, 1, 2, 7, 2, 8, 2, 8, 8, 3, 6, 7, 0, 7, 7, 7, 2, 2, 6, 2, 6, 9, 9, 6, 8, 1, 3, 2, 1, 5, 4, 3, 7, 4, 7, 7, 6, 9, 6, 1, 4, 0, 2, 0, 9, 6, 3, 6, 6, 1, 9, 1, 9, 9, 7, 4, 9, 8, 8, 7, 7, 3, 0, 8, 7, 7, 8, 8, 0, 8
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
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
Cf. A309600, A319740 (10-adic cube root of 1/11).
Digits of 10-adic integers:
A153042 ((-1/9)^(1/3));
A225406 ( 9^(1/3));
A225409 ( (-9)^(1/3)).
Sequence in context: A089065 A248319 A154398 * A176019 A079059 A342574
KEYWORD
nonn,base
AUTHOR
Aswini Vaidyanathan, May 07 2013
EXTENSIONS
Name edited by Seiichi Manyama, Aug 04 2019
STATUS
approved