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

2,2

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).

Links

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

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

Example

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.
...

Prog

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

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)).

Sequence in context: A365492 A024729 A046271 * A283076 A046263 A092279

Adjacent sequences: A341597 A341598 A341599 * A341601 A341602 A341603

Keyword

nonn,easy

Author

Jianing Song, Feb 16 2021

Status

approved