A225405 10-adic integer x such that x^3 = 7.

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

Offset: 0

Author

Aswini Vaidyanathan, May 07 2013

Keywords

• nonn
• base

Examples

			      3^3 == 7 (mod 10).
     43^3 == 7 (mod 10^2).
    543^3 == 7 (mod 10^3).
   1543^3 == 7 (mod 10^4).
  51543^3 == 7 (mod 10^5).
  51543^3 == 7 (mod 10^6).
		

Links

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

Crossrefs

Cf. A225408, A309600.

Programs

PARI

n=0; for(i=1, 100, m=7; for(x=0, 9, if(((n+(x*10^(i-1)))^3)%(10^i)==m, n=n+(x*10^(i-1)); print1(x", "); break)))

PARI

N=100; Vecrev(digits(lift(chinese(Mod((7+O(2^N))^(1/3), 2^N), Mod((7+O(5^N))^(1/3), 5^N)))), N) \\ Seiichi Manyama, Aug 05 2019

Ruby

def A225405(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 7 * (a ** 3 - 7)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A225405(100) # Seiichi Manyama, Aug 13 2019

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 7 * (b(n-1)^3 - 7) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n. - Seiichi Manyama, Aug 13 2019