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a(n) = floor(sqrt(2n-1)).
10

%I #26 Nov 26 2020 22:01:02

%S 1,1,2,2,3,3,3,3,4,4,4,4,5,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7,7,8,8,

%T 8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,11,11,

%U 11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,12,12,12

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

%C n appears 2 * ceiling(n/2) times.

%H Reinhard Zumkeller, <a href="/A103128/b103128.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)+1 for n in A007590, otherwise a(n+1) = a(n).

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

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

%t Table[Floor[Sqrt[2n-1]],{n,100}] (* _Mohammad K. Azarian_, Jun 15 2016 *)

%o (Haskell)

%o a103128 = a000196 . (subtract 1) . (* 2)

%o -- _Reinhard Zumkeller_, Feb 12 2012

%Y Cf. A172471, A000196, A005408, A007590.

%K easy,nonn

%O 1,3

%A _Giovanni Teofilatto_, Mar 17 2005

%E Edited by _Franklin T. Adams-Watters_, Apr 20 2010

%E New name from _Wesley Ivan Hurt_, Nov 26 2020