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%I #73 Sep 08 2022 08:44:39
%S 1,6,20,56,144,352,832,1920,4352,9728,21504,47104,102400,221184,
%T 475136,1015808,2162688,4587520,9699328,20447232,42991616,90177536,
%U 188743680,394264576,822083584,1711276032,3556769792,7381975040,15300820992
%N Expansion of (1+2*x)/(1-2*x)^2.
%C Number of binary trees of size n and height n-1, computed from size n=3 onward; i.e. A014480(n) = A073345(n+3,n+2). (For sizes n=0 through 2 there are no such trees.)
%C Also determinant of the n X n matrix M(i,j)=binomial(2i+2j,i+j). - _Benoit Cloitre_, Mar 27 2004
%C Subdiagonal in triangle displayed in A128196. - _Peter Luschny_, Feb 26 2007
%C From _Jaume Oliver Lafont_, Nov 08 2009: (Start)
%C From two BBP-type formulas by Knuth, (page 6 of the reference)
%C Sum_{n>=0} 1/a(n) = 2^(1/2)*log(1+2^(1/2))
%C Sum_{n>=0} (-1)^n/a(n) = 2^(1/2)*atan(1/2^(1/2))
%C (End)
%C Create a triangle with first column T(n,1)=1+4*n for n=0 1 2... The remaining terms T(r,c)=T(r,c-1)+T(r-1,c-1). T(n,n+1))=a(n). - _J. M. Bergot_, Dec 18 2012
%H Reinhard Zumkeller, <a href="/A014480/b014480.txt">Table of n, a(n) for n = 0..1000</a>
%H David Bailey, Peter Borwein, Simon Plouffe, <a href="https://www.davidhbailey.com/dhbpapers/digits.pdf">On the rapid computation of various polylogarithmic constants</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).
%F a(n) = (2n+1)*2^n = 4a(n-1)-4a(n-2) = 4*A052951(n-1) = a(n-1)+A052951(n) = a(n-1)*(2+4/(2n-1)) = A054582(n, n). - _Henry Bottomley_, May 16 2001
%F E.g.f.: x*cosh(sqrt(2)*x) = x + 6x^3/3! + 20x^5/5! + 56x^7/7! +... - _Ralf Stephan_, Mar 03 2005
%F From _Reinhard Zumkeller_, Apr 27 2006: (Start)
%F a(n) = A118416(n+1,n+1) = A118413(n+1,n+1);
%F A001511(a(n)) = A003602(a(n));
%F A117303(a(n)) = a(n). (End)
%F Row sums of triangle A132775 - _Gary W. Adamson_, Aug 29 2007
%F Row sums of triangle A134233 - _Gary W. Adamson_, Oct 14 2007
%F From _Johannes W. Meijer_, Nov 23 2009: (Start)
%F a(n) = 3*a(n-1) - 2^(n-1)*(2*n-5) with a(0) = 1.
%F a(n) = 3*a(n-1) - 2*a(n-2) + 2^n with a(0) = 1 and a(1) = 6.
%F (End)
%F G.f.: -G(0) where G(k) = 1 - (2*k+2)/(1 - x/(x - (k+1)/G(k+1) )); (continued fraction). - _Sergei N. Gladkovskii_, Dec 06 2012
%F E.g.f.: Q(0), where Q(k)= 1 + 4*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 11 2013
%e (1 + 2*x)/(1-2*x)^2 = 1 + 6*x + 20*x^2 + 56*x^3 + 144*x^4 + 352*x^5 + 832*x^6 + ...
%p a:=n-> sum(2^n*n^binomial(j,n)/2,j=1..n): seq(a(n),n=1..29); # _Zerinvary Lajos_, Apr 18 2009
%t CoefficientList[ Series[(1 + 2*x)/(1 - 2*x)^2, {x, 0, 28}], x]
%t LinearRecurrence[{4, -4}, {1, 6}, 29] (* _Robert G. Wilson v_, Dec 26 2012 *)
%t Table[2^n (2*n + 1), {n, 0, 28}] (* _Fred Daniel Kline_, Oct 20 2014 *)
%o (Haskell)
%o a014480 n = a014480_list !! n
%o a014480_list = 1 : 6 : map (* 4)
%o (zipWith (-) (tail a014480_list) a014480_list)
%o -- _Reinhard Zumkeller_, Jan 22 2012
%o (PARI) Vec((1+2*x)/(1-2*x)^2+O(x^99)) \\ _Charles R Greathouse IV_, Sep 23 2012
%o (Magma) [2^n*(2*n + 1): n in [0..35]]; // _Vincenzo Librandi_, Oct 20 2014
%Y Equals the first left hand column of A167580. - _Johannes W. Meijer_, Nov 12 2009
%Y Cf. A118417, A128196, A132775, A134233.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_.
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