A341602 - OEIS

%I #16 Nov 27 2022 07:58:21

%S 1,0,1,0,0,0,1,1,1,0,1,1,0,0,0,0,0,1,1,0,1,0,0,1,1,1,1,1,1,1,1,0,0,1,

%T 1,0,0,1,1,1,1,0,1,1,1,0,0,0,0,1,0,1,1,1,0,0,0,1,1,0,1,0,0,0,0,1,1,1,

%U 0,1,0,0,0,1,0,1,1,0,0,1,1,1,0,0,1,0,0,0,1,1,0

%N Expansion of the 2-adic integer sqrt(-3/5) that ends in 01.

%C Over the 2-adic integers there are 2 solutions to 5*x^2 + 3 = 0, one ends in 01 and the other ends in 11. This sequence gives the former one. See A341600 for detailed information.

%C This constant may be used to represent one of the two primitive 6th roots of unity, namely one of the two roots of x^2 - x + 1 = 0 in Q_2(sqrt(5)), the unique unramified quatratic extension of the 2-adic field: if x = (1 + A341602*sqrt(5))/2, then x^2 = (-1 + A341602*sqrt(5))/2, x^3 = -1, x^4 = (-1 + A341603*sqrt(5))/2, x^5 = (1 + A341603*sqrt(5))/2 and x^6 = 1.

%C In the ring of 2-adic integers the sequence {Fibonacci(2^(2*n+1))} converges to this constant. For example, Fibonacci(2^21) reduced modulo 2^21 = 1445317 = 101100000110111000101 (binary representation). Reading the binary digits from right to left gives the first 21 terms of this sequence. - _Peter Bala_, Nov 22 2022

%H Jianing Song, <a href="/A341602/b341602.txt">Table of n, a(n) for n = 0..1000</a>

%H Peter Bala, <a href="/A341602/a341602.pdf">Notes on A341602 and A341603</a>

%F a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if 5*A341600(n)^2 + 3 is divisible by 2^(n+2), otherwise 1.

%F a(n) = 1 - A341603(n) for n >= 1.

%F For n >= 2, a(n) = (A341600(n+1) - A341600(n))/2^n.

%e If u = ...11110011001111111100101100000110111000101, then u^2 = ...1001100110011001100110011001100110011001 = -3/5. Furthermore, let x = (1 + u*sqrt(5))/2, then x^2 = (-1 + u*sqrt(5))/2, x^3 = -1, x^4 = (-1 - u*sqrt(5))/2, x^5 = (1 - u*sqrt(5))/2 and x^6 = 1.

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

%Y Cf. A341603, A341600 (successive approximations of the associated 2-adic square root of -3/5), A318962, A318963 (expansion of sqrt(-7)), A322217, A341540 (expansion of sqrt(17)).

%K nonn,base,easy

%O 0

%A _Jianing Song_, Feb 16 2021