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%I #19 Sep 08 2022 08:45:07
%S 0,1,1,2,2,3,4,4,5,5,6,7,7,8,8,9,9,10,11,11,12,12,13,14,14,15,15,16,
%T 16,17,18,18,19,19,20,21,21,22,22,23,24,24,25,25,26,26,27,28,28,29,29,
%U 30,31,31,32,32,33,33,34,35,35,36,36,37,38,38,39,39,40,41,41,42,42,43,43
%N Numerators a(n) of fractions slowly converging to sqrt(2): let a(1) = 0, b(n) = n - a(n); if (a(n) + 1) / b(n) < sqrt(2), then a(n+1) = a(n) + 1, else a(n+1)= a(n).
%C a(n) + b(n) = n and as n -> +infinity, a(n) / b(n) converges to sqrt(2). For all n, a(n) / b(n) < sqrt(2).
%H G. C. Greubel, <a href="/A074840/b074840.txt">Table of n, a(n) for n = 1..5000</a>
%H Heinz H. Bauschke, Minh N. Dao, Scott B. Lindstrom, <a href="https://arxiv.org/abs/1804.08880">The Douglas-Rachford algorithm for a hyperplane and a doubleton</a>, arXiv:1804.08880 [math.OC], 2018.
%F a(1) = 0. b(n) = n - a(n). If (a(n) + 1) / b(n) < sqrt(2), then a(n+1) = a(n) + 1, else a(n+1) = a(n).
%F a(n) = floor(n*(2-sqrt(2))). - _Vladeta Jovovic_, Oct 04 2003
%F a(n) = 2*n - ceiling(n*sqrt(2)). - _Clark Kimberling_, Sep 09 2011
%e a(6)= 3 so b(6) = 6 - 3 = 3. a(7) = 4 because (a(6) + 1) / b(6) = 4/3 which is < sqrt(2). So b(7) = 7 - 4 = 3. a(8) = 4 because (a(7) + 1) / b(7) = 5/3 which is not < sqrt(2).
%t Table[Floor[n*(2 - Sqrt[2])], {n, 1, 50}] (* _G. C. Greubel_, Nov 28 2017 *)
%o (PARI) for(n=1,30, print1(floor(n*(2-sqrt(2))), ", ")) \\ _G. C. Greubel_, Nov 28 2017
%o (Magma) [Floor(n*(2-Sqrt(2))): n in [1..30]]; // _G. C. Greubel_, Nov 28 2017
%Y Cf. A001601.
%K easy,frac,nonn
%O 1,4
%A Robert A. Stump (bee_ess107(AT)msn.com), Sep 09 2002
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