# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a341538 Showing 1-1 of 1 %I A341538 #12 Feb 16 2021 08:39:20 %S A341538 1,1,9,9,41,105,233,233,745,1769,1769,1769,9961,9961,9961,75497, %T A341538 206569,206569,206569,1255145,1255145,5449449,13838057,30615273, %U A341538 64169705,64169705,64169705,332605161,869476073,869476073,869476073,5164443369,13754377961,13754377961 %N A341538 One of the two successive approximations up to 2^n for 2-adic integer sqrt(17). This is the 1 (mod 4) case. %C A341538 a(n) is the unique number k in [1, 2^n] and congruent to 1 mod 4 such that k^2 - 17 is divisible by 2^(n+1). %H A341538 Jianing Song, Table of n, a(n) for n = 2..1000 %F A341538 a(2) = 1; 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 A341538 a(n) = 2^n - A341539(n). %F A341538 a(n) = Sum_{i=0..n-1} A322217(i)*2^i. %e A341538 The unique number k in [1, 4] and congruent to 1 modulo 4 such that k^2 - 17 is divisible by 8 is 1, so a(2) = 1. %e A341538 a(2)^2 - 17 = -16 which is divisible by 16, so a(3) = a(2) = 1. %e A341538 a(3)^2 - 17 = -16 which is not divisible by 32, so a(4) = a(3) + 2^3 = 9. %e A341538 a(4)^2 - 17 = 64 which is divisible by 64, so a(5) = a(4) = 9. %e A341538 a(5)^2 - 17 = 64 which is not divisible by 128, so a(6) = a(5) + 2^5 = 41. %e A341538 ... %o A341538 (PARI) a(n) = truncate(sqrt(17+O(2^(n+1)))) %Y A341538 Cf. A341539 (the 3 (mod 4) case), A322217 (digits of the associated 2-adic square root of 17), A318960, A318961 (successive approximations of sqrt(-7)). %K A341538 nonn %O A341538 2,3 %A A341538 _Jianing Song_, Feb 13 2021 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE