cp's OEIS Frontend

A184674 a(n) = n+floor((n/2-1/(2*n))^2); complement of A184675.

%I A184674 #30 Apr 04 2023 22:17:24
%S A184674 1,2,4,7,10,14,18,23,28,34,40,47,54,62,70,79,88,98,108,119,130,142,
%T A184674 154,167,180,194,208,223,238,254,270,287,304,322,340,359,378,398,418,
%U A184674 439,460,482,504,527,550,574,598,623,648,674,700,727,754,782,810,839,868,898,928,959,990,1022,1054,1087,1120,1154,1188,1223,1258,1294,1330,1367,1404
%N A184674 a(n) = n+floor((n/2-1/(2*n))^2); complement of A184675.
%C A184674 Conjecture: a(n) = A014616(n-1). - _R. J. Mathar_, Jan 29 2011
%C A184674 The above conjecture is true. - _Stefano Spezia_, Apr 04 2023
%H A184674 Vincenzo Librandi, <a href="/A184674/b184674.txt">Table of n, a(n) for n = 1..10000</a>
%H A184674 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F A184674 a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>=6.
%F A184674 G.f.: x*(x^4 - x^3 - 1)/((x + 1)*(x - 1)^3). - _Álvar Ibeas_, Jul 20 2021
%F A184674 a(n) = (2*n^2 + 8*n - 9 + (-1)^n)/8 for n > 1. - _Stefano Spezia_, Apr 04 2023
%p A184674 A184674:=n->n+floor((n/2-1/(2*n))^2): seq(A184674(n), n=1..100); # _Wesley Ivan Hurt_, Feb 22 2017
%t A184674 a[n_]:=n+Floor[(n/2-1/(2n))^2];
%t A184674 b[n_]:=n+Floor[n^(1/2)+(n+1)^(1/2)];
%t A184674 Table[a[n],{n,1,120}]   (* A184674 *)
%t A184674 Table[b[n],{n,1,120}]   (* A184675 *)
%t A184674 FindLinearRecurrence[Table[a[n],{n,1,120}]]
%t A184674 Join[{1},LinearRecurrence[{2,0,-2,1},{2,4,7,10},72]] (* _Ray Chandler_, Aug 02 2015 *)
%o A184674 (Magma) [n+Floor((n/2-1/(2*n))^2): n in [1..80]]; // _Vincenzo Librandi_, Jul 10 2011
%Y A184674 Cf. A184675, A184676, A014616.
%K A184674 nonn,easy
%O A184674 1,2
%A A184674 _Clark Kimberling_, Jan 19 2011