|
%I #25 Aug 23 2018 02:10:56
%S 1,2,5,11,20,32,47,65,86,110,137,167,200,236,275,317,362,410,461,515,
%T 572,632,695,761,830,902,977,1055,1136,1220,1307,1397,1490,1586,1685,
%U 1787,1892,2000,2111,2225
%N Binomial transform of [1, 1, 2, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, ...].
%C A007318 * [1, 1, 2, 1, -1, 1, -1, 1, ...].
%C The quadratic expression for a(n) follows at once by taking into account that the alternate row sums in the Pascal triangle are equal to zero (starting with the second row). - _Emeric Deutsch_, May 03 2008
%C For n > 1, 3*(8*a(n) - 13) = A016945(n-2)^2. - _Vincenzo Librandi_, Feb 15 2012
%H Vincenzo Librandi, <a href="/A139482/b139482.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: (x^3+2*x^2-x+1)/(-x^3+3*x^2-3*x+1). - _Alexander R. Povolotsky_, Apr 24 2008
%F a(n) = (10 - 9*n + 3*n^2)/2 for n >= 2. - _Emeric Deutsch_, May 03 2008
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=2, a(3)=5, a(4)=11. - _Harvey P. Dale_, May 02 2015
%e a(4) = 11 = (1, 3, 3, 1) dot (1, 1, 2, 1) = (1 + 3 + 6 + 1).
%p 1,seq((10+3*n^2-9*n)*1/2,n=2..40); # _Emeric Deutsch_, May 03 2008
%t Join[{1,2},FoldList[##+3&,5,3*Range@100]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 17 2011 *)
%t LinearRecurrence[{3,-3,1},{1,2,5,11},40] (* _Harvey P. Dale_, May 02 2015 *)
%K nonn,easy
%O 1,2
%A _Gary W. Adamson_, Apr 23 2008
%E More terms from _Emeric Deutsch_, May 03 2008
|
