A286841 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 8 (mod 13) case (except for n=0).

0, 8, 99, 1958, 28322, 228249, 2827300, 55922199, 808904403, 9781942334, 52199939826, 603633907222, 11356596271444, 11356596271444, 1828607235824962, 37264994707118563, 651495710876207647, 5974828584341646375, 49226908181248336040

Offset

0,2

Links

Seiichi Manyama, Table of n, a(n) for n = 0..899

Wikipedia, Hensel's Lemma.

Formula

If n > 0, a(n) = 13^n - A286840(n).

a(0) = 0 and a(1) = 8, a(n) = a(n-1) + 4 * (a(n-1)^2 + 1) mod 13^n for n > 1.

a(n) == L(13^n,8) (mod 13^n) == (4 + sqrt(17))^(13^n) + (4 - sqrt(17))^(13^n) (mod 13^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

Mathematica

{0}~Join~Table[#&@@Select[PowerModList[-1, 1/2, 13^k], Mod[#, 13] == 8 &], {k, 18}] (* Giorgos Kalogeropoulos, Oct 22 2022 *)

Prog

(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 A286841(n)
  A(8, 13, n)
end
p A286841(100)
(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 a286841(n):
    return A(8, 13, n)
print(a286841(100)) # Indranil Ghosh, Aug 03 2017, after Ruby
(PARI) a(n) = if (n, 13^n - truncate(sqrt(-1+O(13^n))), 0); \\ Michel Marcus, Aug 04 2017

Crossrefs

The two successive approximations up to p^n for p-adic integer sqrt(-1): A048898 and A048899 (p=5), A286840 and this sequence (p=13), A286877 and A286878 (p=17).

Cf. A114525, A286839.

Sequence in context: A230343 A293145 A305919 * A356440 A367445 A316870

Adjacent sequences: A286838 A286839 A286840 * A286842 A286843 A286844

Keyword

nonn,easy

Author

Seiichi Manyama, Aug 01 2017

Status

approved