A341600 One of the two successive approximations up to 2^n for 2-adic integer sqrt(-3/5). This is the 1 (mod 4) case.

1, 5, 5, 5, 5, 69, 197, 453, 453, 1477, 3525, 3525, 3525, 3525, 3525, 3525, 134597, 396741, 396741, 1445317, 1445317, 1445317, 9833925, 26611141, 60165573, 127274437, 261492165, 529927621, 1066798533, 2140540357, 2140540357, 2140540357, 10730474949, 27910344133

Offset: 2

Author

Jianing Song, Feb 16 2021

Keywords

• nonn
• easy

Comments

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

Examples

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

Links

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

Crossrefs

Cf. A341601 (the 3 (mod 4) case), A341602 (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)).

Programs

PARI

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

Formula

a(2) = 1; 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 - A341601(n).

a(n) = Sum_{i=0..n-1} A341602(i)*2^i.

a(n) == Fibonacci(2^(2*n-1)) (mod 2^n). - Peter Bala, Nov 11 2022