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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.
4
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
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
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Feb 16 2021
STATUS
approved