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A249453 a(0) = 4; for n>0, a(n) = a(n-1) + 2^n - 3.

%I A249453 #32 Sep 08 2022 08:46:10
%S A249453 4,3,4,9,22,51,112,237,490,999,2020,4065,8158,16347,32728,65493,
%T A249453 131026,262095,524236,1048521,2097094,4194243,8388544,16777149,
%U A249453 33554362,67108791,134217652,268435377,536870830,1073741739,2147483560,4294967205,8589934498
%N A249453 a(0) = 4; for n>0, a(n) = a(n-1) + 2^n - 3.
%H A249453 Colin Barker, <a href="/A249453/b249453.txt">Table of n, a(n) for n = 0..1000</a>
%H A249453 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,2).
%F A249453 a(n) = 2*a(n-1) + 3*n - 8.
%F A249453 a(n) = 3*a(n-1) - 2*a(n-2) + 3.
%F A249453 From _Colin Barker_, Oct 30 2014: (Start)
%F A249453 a(n) = 2+2^(1+n)-3*n.
%F A249453 G.f.: -(12*x^2-13*x+4) / ((x-1)^2*(2*x-1)).
%F A249453 (End)
%F A249453 E.g.f.: (2-3*x)*exp(x) + 2*exp(2*x). - _Robert Israel_, Nov 23 2014
%e A249453 Start with a(0)=4 and step = -1, then increase the step after each iteration by 2^n: 4-1 = 3; step+2 = 1; 3+1 = 4; step+4 = 5; 4+5 = 9; step+8 = 13; 9+13 = 22; step+16=29; 22+29 = 51; ... - _M. F. Hasler_, Oct 31 2014
%p A249453 seq(2+2^(1+n)-3*n, n=0..100); # _Robert Israel_, Nov 23 2014
%t A249453 a249453[n_Integer] := Module[{a}, a[0] = 4; a[k_] := a[k - 1] + 2^k - 3; a /@ Range[0, n]]; a249453[32] (* _Michael De Vlieger_, Nov 23 2014 *)
%t A249453 CoefficientList[Series[(12 x^2 - 13 x+4)/((1 - x)^2 (1 - 2 x)), {x, 0, 40}], x] (* _Vincenzo Librandi_, Oct 31 2014 *)
%o A249453 (Magma) [4] cat [n le 1 select 3 else Self(n-1)+2^n-3: n in [1..32]];
%o A249453 (Magma) [2+2^(1+n)-3*n: n in [0..32]]; // _Vincenzo Librandi_, Oct 31 2014
%o A249453 (PARI) Vec(-(12*x^2-13*x+4)/((x-1)^2*(2*x-1)) + O(x^100)) \\ _Colin Barker_, Oct 30 2014
%Y A249453 Cf. A094177, which may be an erroneous version of this sequence.
%K A249453 nonn,easy
%O A249453 0,1
%A A249453 _Arkadiusz Wesolowski_, Oct 29 2014