A014480 Expansion of g.f. (1+2*x)/(1-2*x)^2.

1, 6, 20, 56, 144, 352, 832, 1920, 4352, 9728, 21504, 47104, 102400, 221184, 475136, 1015808, 2162688, 4587520, 9699328, 20447232, 42991616, 90177536, 188743680, 394264576, 822083584, 1711276032, 3556769792, 7381975040, 15300820992, 31675383808, 65498251264

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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.)

Also determinant of the n X n matrix M(i,j)=binomial(2i+2j,i+j). - Benoit Cloitre, Mar 27 2004

Subdiagonal in triangle displayed in A128196. - Peter Luschny, Feb 26 2007

From Jaume Oliver Lafont, Nov 08 2009: (Start)

From two BBP-type formulas by Knuth, (page 6 of the reference)

Sum_{n>=0} 1/a(n) = 2^(1/2)*log(1+2^(1/2))

Sum_{n>=0} (-1)^n/a(n) = 2^(1/2)*atan(1/2^(1/2))

(End)

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

Examples

			(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 + ...
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• David Bailey, Peter Borwein, and Simon Plouffe, On the rapid computation of various polylogarithmic constants, in: L. Berggren, J. Borwein, and P. Borwein (eds.), Pi: A Source Book, Springer, New York, NY, 2000.
• Index entries for linear recurrences with constant coefficients, signature (4,-4).

Crossrefs

Cf. A054582, A118417, A128196, A132775, A134233.

Leftmost column of A167580 (shifted).

Programs

Haskell

a014480 n = a014480_list !! n
a014480_list = 1 : 6 : map (* 4)
   (zipWith (-) (tail a014480_list) a014480_list)
-- Reinhard Zumkeller, Jan 22 2012

Magma

[2^n*(2*n + 1): n in [0..35]]; // Vincenzo Librandi, Oct 20 2014

Maple

a:=n-> sum(2^n*n^binomial(j,n)/2,j=1..n): seq(a(n),n=1..29); # Zerinvary Lajos, Apr 18 2009

Mathematica

CoefficientList[ Series[(1 + 2*x)/(1 - 2*x)^2, {x, 0, 28}], x]
LinearRecurrence[{4, -4}, {1, 6}, 29] (* Robert G. Wilson v, Dec 26 2012 *)
Table[2^n (2*n + 1), {n, 0, 28}] (* Fred Daniel Kline, Oct 20 2014 *)

PARI

Vec((1+2*x)/(1-2*x)^2+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

Formula

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

E.g.f.: x*cosh(sqrt(2)*x) = x + 6x^3/3! + 20x^5/5! + 56x^7/7! +... - Ralf Stephan, Mar 03 2005

From Reinhard Zumkeller, Apr 27 2006: (Start)

a(n) = A118416(n+1,n+1) = A118413(n+1,n+1);

A001511(a(n)) = A003602(a(n));

A117303(a(n)) = a(n). (End)

Row sums of triangle A132775 - Gary W. Adamson, Aug 29 2007

Row sums of triangle A134233 - Gary W. Adamson, Oct 14 2007

From Johannes W. Meijer, Nov 23 2009: (Start)

a(n) = 3*a(n-1) - 2^(n-1)*(2*n-5) with a(0) = 1.

a(n) = 3*a(n-1) - 2*a(n-2) + 2^n with a(0) = 1 and a(1) = 6.

(End)

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

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