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a(n) = 2^(n-1) - n*(n+1)/2.
2

%I #30 Apr 02 2026 14:49:59

%S 0,-1,-2,-2,1,11,36,92,211,457,958,1970,4005,8087,16264,32632,65383,

%T 130901,261954,524078,1048345,2096899,4194028,8388308,16776891,

%U 33554081,67108486,134217322,268435021,536870447,1073741328,2147483120,4294966735,8589933997

%N a(n) = 2^(n-1) - n*(n+1)/2.

%H Vincenzo Librandi, <a href="/A014846/b014846.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,7,-2).

%F a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4); a(1)=0, a(2)=-1, a(3)=a(4)=-2. - _Harvey P. Dale_, Aug 02 2011

%F From _Elmo R. Oliveira_, Apr 01 2026: (Start)

%F O.g.f.: -x^2*(x^2 - 3*x + 1)/((2*x - 1)*(x - 1)^3).

%F E.g.f.: -(1 + x*(x + 2)*exp(x) - exp(2*x))/2. (End)

%t Table[2^n-n*(n+1),{n,50}]/2 (* _Vladimir Joseph Stephan Orlovsky_, Apr 25 2010 *)

%t LinearRecurrence[{5,-9,7,-2},{0,-1,-2,-2},50] (* _Harvey P. Dale_, Aug 02 2011 *)

%o (Magma) [2^(n-1) - n*(n+1)/2: n in [1..35]]; // _Vincenzo Librandi_, Jul 28 2011

%K sign,easy,changed

%O 1,3

%A _N. J. A. Sloane_

%E More terms from _Vladimir Joseph Stephan Orlovsky_, Apr 25 2010