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a(n) = 2 * 3^(n-2)*n*(1+2*n).
2

%I #22 Sep 08 2022 08:45:03

%S 0,2,20,126,648,2970,12636,51030,198288,747954,2755620,9959598,

%T 35429400,124357194,431530092,1482720390,5050815264,17075199330,

%U 57338232372,191385721566,635369601960,2099044209402,6903833113980

%N a(n) = 2 * 3^(n-2)*n*(1+2*n).

%C Define a triangle with left (first) column T(n,0)=n^2 for n=0,1,2,3.. and the remaining terms T(r,c) = T(r-1,c-1) + 2*T(r,c-1). Then T(n,n) = a(n) on the diagonal. T(n,1) = A056105(n). - _J. M. Bergot_, Jan 26 2013

%H Harry J. Smith, <a href="/A062189/b062189.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-27,27).

%F a(n) = A002943(n)*A000244(n-2). Binomial transform of A007758.

%F G.f.: 2*x*(1+x)/(1-3*x)^3. - _Ralf Stephan_, Mar 13 2003

%F a(n) = 2*A077616(n). - _R. J. Mathar_, Jan 29 2013

%F E.g.f.: 2*x*(1+2*x)*exp(3*x). - _G. C. Greubel_, Jun 06 2019

%t Table[2*3^(n-2)*n*(1+2*n), {n,0,30}] (* _G. C. Greubel_, Jun 06 2019 *)

%t LinearRecurrence[{9,-27,27},{0,2,20},30] (* _Harvey P. Dale_, Jun 08 2022 *)

%o (PARI) { for (n=0, 200, write("b062189.txt", n, " ", n*(4*n + 2)*3^(n - 2)) ) } \\ _Harry J. Smith_, Aug 02 2009

%o (Magma) [2*3^(n-2)*n*(1+2*n): n in [0..30]]; // _G. C. Greubel_, Jun 06 2019

%o (Sage) [2*3^(n-2)*n*(1+2*n) for n in (0..30)] # _G. C. Greubel_, Jun 06 2019

%o (GAP) List([0..30], n-> 2*3^(n-2)*n*(1+2*n)) # _G. C. Greubel_, Jun 06 2019

%K nonn,easy

%O 0,2

%A _Henry Bottomley_, Jun 13 2001