A225411 10-adic integer x such that x^3 == (6*(10^(n+1)-1)/9)+1 mod 10^(n+1) for all n.

3, 2, 5, 3, 5, 5, 4, 1, 6, 5, 5, 3, 1, 6, 8, 4, 8, 9, 4, 8, 3, 0, 3, 8, 6, 8, 6, 2, 7, 7, 5, 3, 5, 5, 6, 1, 6, 4, 6, 6, 4, 3, 7, 4, 7, 6, 7, 1, 3, 6, 5, 8, 1, 0, 3, 2, 9, 5, 1, 6, 3, 2, 8, 8, 3, 3, 4, 7, 9, 2, 5, 9, 0, 7, 1, 9, 6, 9, 6, 9, 2, 6, 0, 2, 0, 2, 6, 5, 3, 4, 6, 8, 6, 9, 9, 1, 2, 5, 4, 2

Offset

0,1

Comments

This is the 10's complement of A225402.

Equivalently, the 10-adic cube root of 1/3, i.e., solution to 3*x^3 = 1 (mod 10^n) for all n. - M. F. Hasler, Jan 02 2019

Links

Robert Israel, Table of n, a(n) for n = 0..10000

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 9 * (3 * b(n-1)^3 - 1) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 12 2019

Example

       3^3 ==      7 (mod 10).
      23^3 ==     67 (mod 10^2).
     523^3 ==    667 (mod 10^3).
    3523^3 ==   6667 (mod 10^4).
   53523^3 ==  66667 (mod 10^5).
  553523^3 == 666667 (mod 10^6).

Maple

op([1, 3], padic:-rootp(3*x^3  -1,  10, 101)); # Robert Israel, Aug 04 2019

Prog

(PARI) n=0; for(i=1, 100, m=(2*(10^i-1)/3)+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((m+O(5^N))^m, 5^N), Mod((m+O(2^N))^m, 2^N)))), N) \\ Following Andrew Howroyd's code for A319740. - M. F. Hasler, Jan 02 2019
(Ruby)
def A225411(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 9 * (3 * a ** 3 - 1)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225411(100) # Seiichi Manyama, Aug 12 2019

Crossrefs

Cf. A225402, A309600, A319740 (10-adic cube root of 1/11).

Sequence in context: A206562 A155981 A057958 * A247815 A057953 A372679

Adjacent sequences: A225408 A225409 A225410 * A225412 A225413 A225414

Keyword

nonn,base

Author

Aswini Vaidyanathan, May 07 2013

Status

approved