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a(n) = floor(n/(exp(1/(2*n))-1)).
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%I #18 Nov 23 2023 18:00:02

%S 1,7,16,30,47,69,94,124,157,195,236,282,331,385,442,504,569,639,712,

%T 790,871,957,1046,1140,1237,1339,1444,1554,1667,1785,1906,2032,2161,

%U 2295,2432,2574,2719,2869,3022,3180,3341,3507

%N a(n) = floor(n/(exp(1/(2*n))-1)).

%C Equations of this general form: (n/(exp(1/(r*n))-1)) have a fractional portion that converges to one or more rational fractions if r is rational. They have second differences that are nearly constant before the floor function, and repeat in patterns when calculated after the floor function.

%C The fractional portion of this equation (before the floor function) oscillates between two fractions that converge towards 1/24 and 13/24.

%C Second differences of a(n) = repeat{3,5}.

%C First differences of a(n) = A075123(n+3).

%C Partial sums of a(n) = A033951(n).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).

%F From _Ralf Stephan_, Mar 28 2014: (Start)

%F a(n) = (1/4)*(8n^2 - 2n - 1 + (-1)^n).

%F G.f.: x*(2*x^2 + 5*x + 1)/((1-x^2)*(1-x)^2). (End)

%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - _Wesley Ivan Hurt_, Apr 01 2022

%F E.g.f.: (x*(4*x + 3)*cosh(x) + (4*x^2 + 3*x - 1)*sinh(x))/2. - _Stefano Spezia_, Nov 23 2023

%t Table[Floor[n/(Exp[1/(2 n)] - 1)], {n, 100}] (* _Wesley Ivan Hurt_, Apr 01 2022 *)

%Y Cf. A033951, A075123.

%K nonn,easy

%O 1,2

%A _Richard R. Forberg_, Dec 29 2013