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a(n) = n + floor(sqrt(2n+1)).
1

%I #16 Sep 12 2016 16:06:44

%S 2,4,5,7,8,9,10,12,13,14,15,17,18,19,20,21,22,24,25,26,27,28,29,31,32,

%T 33,34,35,36,37,38,40,41,42,43,44,45,46,47,49,50,51,52,53,54,55,56,57,

%U 58,60,61,62,63,64,65,66,67,68,69,71,72,73,74,75,76,77,78,79,80,81,82,84,85,86,87,88,89,90,91,92

%N a(n) = n + floor(sqrt(2n+1)).

%H Robert Israel, <a href="/A183573/b183573.txt">Table of n, a(n) for n = 1..10000</a>

%F From _Robert Israel_, Sep 12 2016: (Start)

%F a(n+1)=a(n)+2 if n is in A047838, otherwise a(n+1) = a(n)+1.

%F a(n) = n + A103128(n+1).

%F G.f.: Theta3(x^2)/(2*(1-x)) + Theta2(x^2)/(2*sqrt(x)*(1-x)) - (1-2*x)*(3-x)/(2*(1-x)^2), where Theta2 and Theta3 are Jacobi Theta functions. (End)

%p seq(n+floor(sqrt(2*n+1)), n=1..100); # _Robert Israel_, Sep 12 2016

%t Table[n + Floor@ Sqrt[2 n + 1], {n, 80}] (* _Michael De Vlieger_, Sep 12 2016 *)

%Y Cf. A047838, A103128, A116940 (complement).

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 05 2011