A341539 One of the two successive approximations up to 2^n for 2-adic integer sqrt(17). This is the 3 (mod 4) case.

3, 7, 7, 23, 23, 23, 23, 279, 279, 279, 2327, 6423, 6423, 22807, 55575, 55575, 55575, 317719, 842007, 842007, 2939159, 2939159, 2939159, 2939159, 2939159, 70048023, 204265751, 204265751, 204265751, 1278007575, 3425491223, 3425491223, 3425491223, 20605360407

Offset: 2

Author

Jianing Song, Feb 13 2021

Keywords

• nonn

Comments

a(n) is the unique number k in [1, 2^n] and congruent to 3 mod 4 such that k^2 - 17 is divisible by 2^(n+1).

Examples

			The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^2 - 17 is divisible by 8 is 3, so a(2) = 3.
a(2)^2 - 17 = -8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 7.
a(3)^2 - 17 = 32 which is divisible by 32, so a(4) = a(3) = 7.
a(4)^2 - 17 = 32 which is not divisible by 64, so a(5) = a(4) + 2^4 = 23.
a(5)^2 - 17 = 512 which is divisible by 128, so a(6) = a(5) = 23.
...
		

Links

• Jianing Song, Table of n, a(n) for n = 2..1000

Crossrefs

Cf. A341538 (the 1 (mod 4) case), A341540 (digits of the associated 2-adic square root of 17), A318960, A318961 (successive approximations of sqrt(-7)).

Programs

Mathematica

Table[First@Select[PowerModList[17,1/2,2^(k+1)],Mod[#,4]==3&],{k,2,35}] (* Giorgos Kalogeropoulos, Oct 22 2022 *)

PARI

a(n) = if(n==2, 3, truncate(-sqrt(17+O(2^(n+1)))))

Formula

a(2) = 3; for n >= 3, a(n) = a(n-1) if a(n-1)^2 - 17 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).

a(n) = 2^n - A341538(n).

a(n) = Sum_{i=0..n-1} A341540(i)*2^i.