OFFSET
0
COMMENTS
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.
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.
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
LINKS
Jianing Song, Table of n, a(n) for n = 0..1000
Peter Bala, Notes on A341602 and A341603
FORMULA
EXAMPLE
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.
PROG
(PARI) a(n) = truncate(-sqrt(-3/5+O(2^(n+2))))\2^n
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Jianing Song, Feb 16 2021
STATUS
approved