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a(n) = n^n*binomial(n+2, 2).
7

%I #19 Sep 08 2022 08:45:09

%S 1,3,24,270,3840,65625,1306368,29647548,754974720,21308126895,

%T 660000000000,22254310307658,811365140791296,31801886192186565,

%U 1333440819066961920,59553569091796875000,2822351843277561397248

%N a(n) = n^n*binomial(n+2, 2).

%C A diagonal of A081130.

%H Vincenzo Librandi, <a href="/A081133/b081133.txt">Table of n, a(n) for n = 0..100</a>

%F a(n) = n^n*(n+1)*(n+2)/2.

%p seq(n^n*binomial(n+2,2), n=0..20); # _G. C. Greubel_, May 18 2021

%t Join[{1},Table[n^n Binomial[n+2,2],{n,20}]] (* _Harvey P. Dale_, Dec 27 2011 *)

%o (Magma) [(n^n*Binomial(n+2,2)): n in [0..20]]; // _Vincenzo Librandi_, Sep 22 2011

%o (Sage) [n^n*binomial(n+2,2) for n in (0..20)] # _G. C. Greubel_, May 18 2021

%Y Sequences of the form (n+m)^n*binomial(n+2,2): this sequence (m=0), A081132 (m=1), A081131 (m=2), A053507 (m=3), A081196 (m=4).

%K nonn,easy

%O 0,2

%A _Paul Barry_, Mar 08 2003