A341601 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-3/5). This is the 3 (mod 4) case.

3, 3, 11, 27, 59, 59, 59, 59, 571, 571, 571, 4667, 12859, 29243, 62011, 127547, 127547, 127547, 651835, 651835, 2748987, 6943291, 6943291, 6943291, 6943291, 6943291, 6943291, 6943291, 6943291, 6943291, 2154426939, 6449394235, 6449394235, 6449394235

Offset

2,1

Comments

a(n) is the unique number k in [1, 2^n] and congruent to 3 mod 4 such that 5*k^2 + 3 is divisible by 2^(n+1).

Links

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

Formula

a(2) = 3; for n >= 3, a(n) = a(n-1) if 5*a(n-1)^2 + 3 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).

a(n) = 2^n - A341600(n).

a(n) = Sum_{i = 0..n-1} A341603(i)*2^i.

a(n) == Fibonacci(4^n) (mod 2^n). - Peter Bala, Nov 11 2022

Example

The unique number k in [1, 4] and congruent to 3 modulo 4 such that 5*k^2 + 3 is divisible by 8 is 3, so a(2) = 3.
5*a(2)^2 + 3 = 48 which is divisible by 16, so a(3) = a(2) = 3.
5*a(3)^2 + 3 = 48 which is not divisible by 32, so a(4) = a(3) + 2^3 = 11.
5*a(4)^2 + 3 = 608 which is not divisible by 64, so a(5) = a(4) + 2^4 = 27.
5*a(5)^2 + 3 = 3648 which is not divisible by 128, so a(6) = a(5) + 2^5 = 59.
...

Prog

(PARI) a(n) = if(n==2, 3, truncate(sqrt(-3/5+O(2^(n+1)))))

Crossrefs

Cf. A145231, A341600 (the 1 (mod 4) case), A341603 (digits of the associated 2-adic square root of -3/5), A318960, A318961 (successive approximations of sqrt(-7)), A341538, A341539 (successive approximations of sqrt(17)).

Sequence in context: A281284 A281639 A281101 * A322701 A124265 A163938

Adjacent sequences: A341598 A341599 A341600 * A341602 A341603 A341604

Keyword

nonn,easy

Author

Jianing Song, Feb 16 2021

Status

approved