A225402 10-adic cube root of -1/3.

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

Offset: 0

Author

Aswini Vaidyanathan, May 07 2013

Keywords

• nonn
• base

Comments

Previous name was: 10-adic integer x such that x^3 == (10^(n+1)-1)/3 mod 10^(n+1) for n >= 0.

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

Examples

			       7^3 == 3      (mod 10).
      77^3 == 33     (mod 10^2).
     477^3 == 333    (mod 10^3).
    6477^3 == 3333   (mod 10^4).
   46477^3 == 33333  (mod 10^5).
  446477^3 == 333333 (mod 10^6).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A225411, A309600, A319740 (10-adic cube root of 1/11).

Programs

Maple

b:= proc(n) option remember; `if`(n<2, 7*n,
      irem(b(n-1)+9*(3*b(n-1)^3+1), 10^n))
    end:
a:= n-> (b(n+1)-b(n))/10^n:
seq(a(n), n=0..100);  # Alois P. Heinz, Apr 15 2022

PARI

n=0;for(i=1,100,m=(10^i-1)/3;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 A225402(n)
  ary = [7]
  a = 7
  n.times{|i|
    b = (a + 9 * (3 * a ** 3 + 1)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225402(100) # Seiichi Manyama, Aug 12 2019

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 7, 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

Extensions

New name (using M. F. Hasler's comment) from Joerg Arndt, Aug 08 2019