A286877 One of the two successive approximations up to 17^n for 17-adic integer sqrt(-1). Here the 4 (mod 17) case (except for n=0).

0, 4, 38, 2928, 27493, 1029745, 23747457, 313398285, 3596107669, 94280954402, 450044583893, 28673959190179, 28673959190179, 3524407382568745, 13428985415474682, 13428985415474682, 42949774758062711577, 91610966633729580058, 6709533061724423693474

Offset: 0

Author

Seiichi Manyama, Aug 02 2017

Keywords

• nonn
• easy

Comments

x = ...GC5A24,

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

Examples

			a(1) = (   4)_17 = 4,
a(2) = (  24)_17 = 38,
a(3) = ( A24)_17 = 2928,
a(4) = (5A24)_17 = 27493.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..813
• Peter Bala, Using Lucas polynomials to find the p-adic square roots of -1, -2 and -3, Dec 2022.
• 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), this sequence and A286878 (p=17).

Programs

PARI

a(n) = truncate(sqrt(-1+O(17^n))); \\ 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 a286877(n):
    return A(4, 17, n)
print(a286877(100)) # Indranil Ghosh, Aug 03 2017

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 A286877(n)
  A(4, 17, n)
end
p A286877(100)

Formula

a(0) = 0 and a(1) = 4, a(n) = a(n-1) + 2 * (a(n-1)^2 + 1) mod 17^n for n > 1.

a(n) == L(17^n,4) (mod 17^n) == (2 + sqrt(5))^(17^n) + (2 - sqrt(5))^(17^n) (mod 17^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Dec 02 2022