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
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..899
- Wikipedia, Hensel's Lemma.
Crossrefs
Programs
-
Mathematica
{0}~Join~Table[#&@@Select[PowerModList[-1, 1/2, 13^k], Mod[#, 13] == 8 &], {k, 18}] (* Giorgos Kalogeropoulos, Oct 22 2022 *)
-
PARI
a(n) = if (n, 13^n - truncate(sqrt(-1+O(13^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 a286841(n): return A(8, 13, n) print(a286841(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 A286841(n) A(8, 13, n) end p A286841(100)
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