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%I #35 Feb 23 2023 03:46:21
%S 4,3,7,6,10,9,13,12,16,15,19,18,22,21,25,24,28,27,31,30,34,33,37,36,
%T 40,39,43,42,46,45,49,48,52,51,55,54,58,57,61,60,64,63,67,66,70,69,73,
%U 72,76,75,79,78,82,81,85,84,88,87,91,90,94,93,97,96,100,99,103,102,106,105
%N a(n) = 3*n - a(n-1) + 1, with a(1)=4.
%H Vincenzo Librandi, <a href="/A168200/b168200.txt">Table of n, a(n) for n = 1..1100</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F a(n) = (6*n + 5 - 5*(-1)^n)/4. - _Jon E. Schoenfield_, Jun 24 2010
%F From _Joerg Arndt_, Apr 24 2011: (Start)
%F a(n) = +1*a(n-1) + 1*a(n-2) - 1*a(n-3).
%F G.f.: x*(4-x)/(1-x-x^2+x^3) = x*(4-x)/((1+x)*(1-x)^2). (End)
%F a(n) = floor(3*(n+1)/2)-(-1)^n. - _Wesley Ivan Hurt_, Sep 12 2017
%F Sum_{n>=1} (-1)^n/a(n) = 1 - Pi/(6*sqrt(3)) - log(3)/2. - _Amiram Eldar_, Feb 23 2023
%t RecurrenceTable[{a[1]==4,a[n]==3n-a[n-1]+1},a,{n,70}] (* or *) LinearRecurrence[{1,1,-1},{4,3,7},80] (* _Harvey P. Dale_, Jul 31 2014 *)
%o (Magma) [(6*n+5-5*(-1)^n)/4: n in [1..70]];
%o (PARI) a(n)=(6*n+5-5*(-1)^n)/4 \\ _Charles R Greathouse IV_, Jan 11 2012
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Nov 20 2009
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