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A014833 a(n) = 2^n - n*(n+1)/2.

%I A014833 #28 Mar 06 2025 20:21:37
%S A014833 1,1,1,2,6,17,43,100,220,467,969,1982,4018,8101,16279,32648,65400,
%T A014833 130919,261973,524098,1048366,2096921,4194051,8388332,16776916,
%U A014833 33554107,67108513,134217350,268435050,536870477,1073741359,2147483152,4294966768,8589934031,17179868589
%N A014833 a(n) = 2^n - n*(n+1)/2.
%H A014833 Vincenzo Librandi, <a href="/A014833/b014833.txt">Table of n, a(n) for n = 0..240</a>
%H A014833 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,7,-2).
%F A014833 From _Harvey P. Dale_, May 12 2011: (Start)
%F A014833 a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4); a(0)=1, a(1)=1, a(2)=1, a(3)=2.
%F A014833 G.f.: 1/(1-2*x) + 1/(-1+x)^3 + 1/(-1+x)^2. (End)
%F A014833 a(n) = A331105(n,n) for n>0. - _Alois P. Heinz_, Jan 16 2020
%F A014833 E.g.f.: exp(x)*(exp(x) - x*(x + 2)/2). - _Elmo R. Oliveira_, Mar 06 2025
%p A014833 seq(2^n-n*(n+1)/2, n=0..30); # _Zerinvary Lajos_, Jul 01 2007
%t A014833 Table[2^n-n (n+1)/2,{n,0,50}] (* or *) LinearRecurrence[{5,-9,7,-2},{1,1,1,2},50] (* _Harvey P. Dale_, May 12 2011 *)
%o A014833 (Magma) [2^n - n*(n+1)/2: n in [0..50]]; // _Vincenzo Librandi_, Apr 25 2011
%Y A014833 Cf. A000079, A000217, A331105.
%K A014833 nonn,easy
%O A014833 0,4
%A A014833 _N. J. A. Sloane_