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

Author

Aswini Vaidyanathan, May 07 2013

Keywords

• nonn
• base

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

Examples

			       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).
		

Links

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

Crossrefs

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

Programs

Maple

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

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

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