A286878 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 13 (mod 17) case (except for n=0).

0, 13, 251, 1985, 56028, 390112, 390112, 96940388, 3379649772, 24306922095, 1565949316556, 5597937117454, 553948278039582, 6380170650337192, 154948841143926247, 2848994066094341111, 5711417117604156904, 735629295252607184119, 7353551390343301297535

Offset: 0

Author

Seiichi Manyama, Aug 02 2017

Keywords

• nonn

Comments

x = ...04B6ED,

x^2 = ...GGGGGG = -1.

Examples

			a(1) = (   D)_17 = 13,
a(2) = (  ED)_17 = 251,
a(3) = ( 6ED)_17 = 1985,
a(4) = (B6ED)_17 = 56028.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..812
• Wikipedia, Hensel's Lemma.

Crossrefs

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), A286840 and A286841 (p=13), A286877 and this sequence (p=17).

Programs

PARI

a(n) = if (n, 17^n-truncate(sqrt(-1+O(17^n))), 0); \\ Michel Marcus, Aug 04 2017

Python

def A(k, m, n):
      ary=[0]
      a, mod = k, m
      for i in range(n):
          b=a%mod
          ary.append(b)
          a=b**m
          mod*=m
      return ary
def a286878(n): return A(13, 17, n)
print(a286878(100)) # Indranil Ghosh, Aug 03 2017, after Ruby

Ruby

def A(k, m, n)
  ary = [0]
  a, mod = k, m
  n.times{
    b = a % mod
    ary << b
    a = b ** m
    mod *= m
  }
  ary
end
def A286878(n)
  A(13, 17, n)
end
p A286878(100)

Formula

If n > 0, a(n) = 17^n - A286877(n).

a(0) = 0 and a(1) = 13, a(n) = a(n-1) + 15 * (a(n-1)^2 + 1) mod 17^n for n > 1.