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A341603
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Expansion of the 2-adic integer sqrt(-3/5) that ends in 11.
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4
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1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1
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OFFSET
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0
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COMMENTS
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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 latter one. See A341601 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 + A341603*sqrt(5))/2, then x^2 = (-1 + A341603*sqrt(5))/2, x^3 = -1, x^4 = (-1 + A341602*sqrt(5))/2, x^5 = (1 + A341602*sqrt(5))/2 and x^6 = 1.
In the ring of 2-adic integers the sequence {Fibonacci(4^n)} converges to this constant. For example, Fibonacci(4^10) reduced modulo 4^10 = 651835 = 10011111001000111011 (binary representation). Reading the binary digits from right to left gives the first 20 terms of this sequence. - Peter Bala, Nov 22 2022
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LINKS
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FORMULA
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a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if 5*A341601(n)^2 + 3 is divisible by 2^(n+2), otherwise 1.
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EXAMPLE
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If v = ...00001100110000000011010011111001000111011, then v^2 = ...1001100110011001100110011001100110011001 = -3/5. Furthermore, let x = (1 + v*sqrt(5))/2, then x^2 = (-1 + v*sqrt(5))/2, x^3 = -1, x^4 = (-1 - v*sqrt(5))/2, x^5 = (1 - v*sqrt(5))/2 and x^6 = 1.
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PROG
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(PARI) a(n) = if(n==1, 1, truncate(sqrt(-3/5+O(2^(n+2))))\2^n)
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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