login
a(n) = n - 1 + ceiling((4+n^2)/5); complement of A184012.
1

%I #17 Feb 20 2025 12:07:32

%S 1,3,5,7,10,13,17,21,25,30,35,41,47,53,60,67,75,83,91,100,109,119,129,

%T 139,150,161,173,185,197,210,223,237,251,265,280,295,311,327,343,360,

%U 377,395,413,431,450,469,489,509,529,550,571,593,615,637,660,683,707,731,755,780,805,831,857,883,910,937,965,993,1021,1050,1079,1109,1139,1169,1200,1231,1263,1295,1327,1360

%N a(n) = n - 1 + ceiling((4+n^2)/5); complement of A184012.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,1,-2,1).

%F a(n) = n - 1 + ceiling((4+n^2)/5).

%F a(n) = +2 a(n-1) -a(n-2) +a(n-5) -2 a(n-6) +a(n-7). - _R. J. Mathar_, Mar 11 2012

%t a=5; b=-4;

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

%t Table[n-1+Ceiling[(n*n-b)/a],{n,80}]

%t LinearRecurrence[{2,-1,0,0,1,-2,1},{1,3,5,7,10,13,17},80] (* _Harvey P. Dale_, Mar 17 2023 *)

%Y Cf. A184012.

%K nonn,easy,changed

%O 1,2

%A _Clark Kimberling_, Jan 08 2011