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Expansion of the 2-adic integer sqrt(17) that ends in 01.
7

%I #35 Sep 13 2023 12:15:19

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

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

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

%N Expansion of the 2-adic integer sqrt(17) that ends in 01.

%C Over the 2-adic integers there are 2 solutions to x^2 = 17, one ends in 01 and the other ends in 11. This sequence gives the former one. See A341538 for detailed information. - _Jianing Song_, Feb 13 2021

%H Jianing Song, <a href="/A322217/b322217.txt">Table of n, a(n) for n = 0..1000</a>

%F From _Jianing Song_, Feb 14 2021: (Start)

%F a(0) = 1, a(1) = 0; for n >= 2, a(n) = 0 if A341538(n)^2 - 17 is divisible by 2^(n+2), otherwise 1.

%F a(n) = 1 - A341540(n) for n >= 1.

%F For n >= 2, a(n) = (A341538(n+1) - A341538(n))/2^n. (End)

%e sqrt(17) (2-adic) = ...110100010010000011111001100110011110100110010011011101001 = 1 + 2^3 + 2^5 + 2^6 + 2^7 + 2^9 + 2^10 + 2^13 + 2^16 + 2^17 + 2^20 + 2^22 + 2^23 + 2^24 + 2^25 + 2^28 + 2^29 + 2^32 + 2^33 + 2^36 + 2^37 + 2^38 + 2^39 + 2^40 + 2^46 + 2^49 + 2^53 + 2^55 + 2^56 + 2^58 + 2^63 + 2^66 + 2^67 + 2^68 + 2^74 + 2^75 + 2^77 + 2^79 + 2^80 + 2^83 + 2^86 + 2^94 + 2^96 + 2^98 + 2^99 + ...

%e The exponents of 2 in the above series begins:

%e [0, 3, 5, 6, 7, 9, 10, 13, 16, 17, 20, 22, 23, 24, 25, 28, 29, 32, 33, 36, 37, 38, 39, 40, 46, 49, 53, 55, 56, 58, 63, 66, 67, 68, 74, 75, 77, 79, 80, 83, 86, 94, 96, 98, 99, 101, 104, 105, 110, 112, 114, 115, 117, 119, 120, 122, 123, 125, 129, 130, 132, 136, 137, 138, 140, 143, 144, 145, 147, 150, 154, 158, 159, 160, 162, 163, 164, 165, 167, 168, 170, 174, 175, 180, 184, 185, 188, 189, 190, 192, 196, 198, ...].

%e The least positive real solution to the following power series in x with the same exponents:

%e sqrt(17) = 1 + x^3 + x^5 + x^6 + x^7 + x^9 + x^10 + x^13 + x^16 + x^17 + x^20 + x^22 + x^23 + x^24 + x^25 + x^28 + x^29 + ...

%e is x = 0.868455038691709167...

%o (PARI) /* Print the coefficients of powers of 2 in reverse order: */

%o Vecrev(binary(truncate( sqrt(17 + O(2^1001)) )))

%o (PARI) a(n) = truncate(sqrt(17+O(2^(n+2))))\2^n \\ _Jianing Song_, Feb 13 2021

%Y Cf. A341540, A341538 (successive approximations of the associated 2-adic square root of 17), A318962, A318963 (expansion of sqrt(-7)).

%Y Equals A341753^2 = A341754^2.

%K nonn,base

%O 0

%A _Paul D. Hanna_, Dec 14 2018

%E Name edited by _Jianing Song_, Feb 13 2021