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A341539 One of the two successive approximations up to 2^n for 2-adic integer sqrt(17). This is the 3 (mod 4) case.

%I A341539 #13 Dec 02 2022 13:21:00
%S A341539 3,7,7,23,23,23,23,279,279,279,2327,6423,6423,22807,55575,55575,55575,
%T A341539 317719,842007,842007,2939159,2939159,2939159,2939159,2939159,
%U A341539 70048023,204265751,204265751,204265751,1278007575,3425491223,3425491223,3425491223,20605360407
%N A341539 One of the two successive approximations up to 2^n for 2-adic integer sqrt(17). This is the 3 (mod 4) case.
%C A341539 a(n) is the unique number k in [1, 2^n] and congruent to 3 mod 4 such that k^2 - 17 is divisible by 2^(n+1).
%H A341539 Jianing Song, <a href="/A341539/b341539.txt">Table of n, a(n) for n = 2..1000</a>
%F A341539 a(2) = 3; for n >= 3, a(n) = a(n-1) if a(n-1)^2 - 17 is divisible by 2^(n+1), otherwise a(n-1) + 2^(n-1).
%F A341539 a(n) = 2^n - A341538(n).
%F A341539 a(n) = Sum_{i=0..n-1} A341540(i)*2^i.
%e A341539 The unique number k in [1, 4] and congruent to 3 modulo 4 such that k^2 - 17 is divisible by 8 is 3, so a(2) = 3.
%e A341539 a(2)^2 - 17 = -8 which is not divisible by 16, so a(3) = a(2) + 2^2 = 7.
%e A341539 a(3)^2 - 17 = 32 which is divisible by 32, so a(4) = a(3) = 7.
%e A341539 a(4)^2 - 17 = 32 which is not divisible by 64, so a(5) = a(4) + 2^4 = 23.
%e A341539 a(5)^2 - 17 = 512 which is divisible by 128, so a(6) = a(5) = 23.
%e A341539 ...
%t A341539 Table[First@Select[PowerModList[17,1/2,2^(k+1)],Mod[#,4]==3&],{k,2,35}] (* _Giorgos Kalogeropoulos_, Oct 22 2022 *)
%o A341539 (PARI) a(n) = if(n==2, 3, truncate(-sqrt(17+O(2^(n+1)))))
%Y A341539 Cf. A341538 (the 1 (mod 4) case), A341540 (digits of the associated 2-adic square root of 17), A318960, A318961 (successive approximations of sqrt(-7)).
%K A341539 nonn
%O A341539 2,1
%A A341539 _Jianing Song_, Feb 13 2021