A286840 One of the two successive approximations up to 13^n for 13-adic integer sqrt(-1). Here the 5 (mod 13) case (except for n=0).

0, 5, 70, 239, 239, 143044, 1999509, 6826318, 6826318, 822557039, 85658552023, 1188526486815, 11941488851037, 291518510320809, 2108769149874327, 13920898306972194, 13920898306972194, 2675587335039691558, 63228498770709057089, 513050126578538629605

Offset

0,2

Links

Seiichi Manyama, Table of n, a(n) for n = 0..897

Wikipedia, Hensel's Lemma.

Formula

a(0) = 0 and a(1) = 5, a(n) = a(n-1) + 9 * (a(n-1)^2 + 1) mod 13^n for n > 1.

a(n) == L(13^n,5) (mod 13^n) == ((5 + sqrt(29))/2)^(13^n) + ((5 - sqrt(29))/2)^(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] == 5 &], {k, 20}]  (* Giorgos Kalogeropoulos, Oct 21 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 A286840(n)
  A(5, 13, n)
end
p A286840(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 a286840(n):
    return A(5, 13, n)
print(a286840(100)) # Indranil Ghosh, Aug 03 2017, after Ruby
(PARI) a(n) = truncate(sqrt(-1+O(13^n))); \\ 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), this sequence and A286841 (p=13), A286877 and A286878 (p=17).

Cf. A034944, A114525, A286838.

Sequence in context: A299318 A051538 A218709 * A034944 A064046 A256235

Adjacent sequences: A286837 A286838 A286839 * A286841 A286842 A286843

Keyword

nonn,easy

Author

Seiichi Manyama, Aug 01 2017

Status

approved