A183867 - OEIS

%I #22 Sep 08 2022 08:45:55

%S 3,4,6,8,9,10,12,13,15,16,17,18,20,21,22,24,25,26,27,28,30,31,32,33,

%T 35,36,37,38,39,40,42,43,44,45,46,48,49,50,51,52,53,54,56,57,58,59,60,

%U 61,63,64,65,66,67,68,69,70,72,73,74,75,76,77,78,80,81,82,83

%N a(n) = n + floor(2*sqrt(n)); complement of A184676.

%C Also equals n + floor(sqrt(n) + sqrt(n+1/2)). Proof: floor(2*sqrt(n)) is the largest k such that k^2/4 <= n, while floor(sqrt(n) + sqrt(n+1/2)) is the largest k such that (k^2 - 1)/4 + 1/(16*k^2) <= n. All perfect squares are 0 or 1 (mod 4). In either case, it is easily verified that one of the inequalities is satisfied if and only if the other inequality is satisfied. - _Nathaniel Johnston_, Jun 26 2011

%H Nathaniel Johnston, <a href="/A183867/b183867.txt">Table of n, a(n) for n = 1..5000</a>

%p seq(n+floor(2*sqrt(n)), n=1..67); # _Nathaniel Johnston_, Jun 26 2011

%t a=4; b=0;

%t Table[n+Floor[(a*n+b)^(1/2)],{n,100}]

%t Table[n-1+Ceiling[(n*n-b)/a],{n,70}]

%o (PARI) a(n) = n+floor(2*sqrt(n)); \\ _Michel Marcus_, Dec 08 2015

%o (Magma) [n+Floor(2*Sqrt(n)): n in [1..100]]; // _Vincenzo Librandi_, Dec 09 2015

%Y Cf. A179272.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 07 2011