A309613 Digits of the 10-adic integer (13/9)^(1/3).

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

Offset: 0

Author

Seiichi Manyama, Aug 10 2019

Keywords

• nonn
• base

Examples

			       3^3 == 7      (mod 10).
      93^3 == 57     (mod 10^2).
     693^3 == 557    (mod 10^3).
    9693^3 == 5557   (mod 10^4).
   69693^3 == 55557  (mod 10^5).
  669693^3 == 555557 (mod 10^6).
		

Links

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

Crossrefs

Cf. A178769, A309600, A309608.

Programs

PARI

N=100; Vecrev(digits(lift(chinese(Mod((13/9+O(2^N))^(1/3), 2^N), Mod((13/9+O(5^N))^(1/3), 5^N)))), N)

Ruby

def A309613(n)
  ary = [3]
  a = 3
  n.times{|i|
    b = (a + 3 * (9 * a ** 3 - 13)) % (10 ** (i + 2))
    ary << (b - a) / (10 ** (i + 1))
    a = b
  }
  ary
end
p A309613(100)

Formula

Define the sequence {b(n)} by the recurrence b(0) = 0 and b(1) = 3, b(n) = b(n-1) + 3 * (9 * b(n-1)^3 - 13) mod 10^n for n > 1, then a(n) = (b(n+1) - b(n))/10^n.