login
Expansion of ((1+x)/(1-3*x))^2.
3

%I #13 Sep 08 2022 08:45:23

%S 1,8,40,168,648,2376,8424,29160,99144,332424,1102248,3621672,11809800,

%T 38263752,123294312,395392104,1262703816,4017693960,12741829416,

%U 40291730856,127073920392,399817944648,1255242384360,3933092804328

%N Expansion of ((1+x)/(1-3*x))^2.

%C Binomial transform is A014916. 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.

%H G. C. Greubel, <a href="/A113071/b113071.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6,-9).

%F G.f.: (1+x)^2/(1-3*x)^2. [Corrected by _Georg Fischer_, May 24 2019]

%F a(n) = (8*3^n*(2*n+1) + 0^n)/9 = (4*3^n*(4*n+2) + 0^n)/9;

%F a(n) = Sum_{k=0..n} A003946(k)*A003946(n-k).

%F a(n) = Sum_{k=0..n} C(n, k)*Sum_{j=0..k} (j+1)*4^j.

%F a(n) = 8*A081038(n-1), n>0. - _R. J. Mathar_, Nov 28 2014

%t CoefficientList[Series[(1+x)^2/(1-3x)^2, {x, 0, 30}], x] (* _Georg Fischer_, May 24 2019 *)

%t LinearRecurrence[{6,-9}, {1,8,40}, 30] (* _G. C. Greubel_, May 24 2019 *)

%o (PARI) my(x='x+O('x^30)); Vec(((1+x)/(1-3*x))^2) \\ _G. C. Greubel_, May 24 2019

%o (Magma) I:=[8,40]; [1] cat [n le 2 select I[n] else 6*Self(n-1) - 9*Self(n-2): n in [1..30]]; // _G. C. Greubel_, May 24 2019

%o (Sage) (((1+x)/(1-3*x))^2).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, May 24 2019

%o (GAP) a:=[1,8,40];; for n in [4..30] do a[n]:=6*a[n-1]-9*a[n-2]; od; a; # _G. C. Greubel_, May 24 2019

%K easy,nonn

%O 0,2

%A _Paul Barry_, Oct 14 2005