A113070 Expansion of ((1+x)/(1-2x))^2.

1, 6, 21, 60, 156, 384, 912, 2112, 4800, 10752, 23808, 52224, 113664, 245760, 528384, 1130496, 2408448, 5111808, 10813440, 22806528, 47972352, 100663296, 210763776, 440401920, 918552576, 1912602624, 3976200192, 8254390272, 17112760320

Offset: 0

Author

Paul Barry, Oct 14 2005

Keywords

• easy
• nonn

Comments

Binomial transform is A014915. In general, ((1+x)/(1-r*x))^2 expands to a(n) = ((r+1)*r^n*((r+1)*n + r - 1) + 0^n)/r^2, which is also a(n) = Sum_{k=0..n} C(n,k)*Sum_{j=0..k} (j+1)*(r+1)^j. This is the self-convolution of the coordination sequence for the infinite tree with valency r.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-4).

Crossrefs

Cf. A113071.

Programs

Magma

[3*2^n*(3*n+1)/4+0^n/4: n in [0..30]]; // Vincenzo Librandi, May 21 2011

Mathematica

Join[{1},LinearRecurrence[{4,-4},{6,21},30]] (* or *) CoefficientList[ Series[((1+x)/(1-2x))^2,{x,0,30}],x] (* Harvey P. Dale, May 20 2011 *)

Formula

G.f.: (1+x)^2/(1-2x)^2;

a(n) = 3*2^n(3n+1)/4 + 0^n/4;

a(n) = Sum_{k=0..n} A003945(k)*A003945(n-k);

a(n) = Sum_{k=0..n} C(n, k)*Sum_{j=0..k} (j+1)*3^j.

a(n) = 4*a(n-1) - 4*a(n-2); a(0)=1, a(1)=6, a(2)=21. - Harvey P. Dale, May 20 2011