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a(0) = 1; a(n) = n^(n-1)(3n-1)/2 (n>0)
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%I #4 Mar 30 2012 18:58:49

%S 1,1,5,36,352,4375,66096,1176490,24117248,559607373,14500000000,

%T 414998793616,13002646487040,442663617327139,16271152851709952,

%U 642244372558593750,27093655358260903936,1216529796891671712025

%N a(0) = 1; a(n) = n^(n-1)(3n-1)/2 (n>0)

%C Main diagonal of square array T(n,k) where T(n,k)=k^n(n^2-n+2k^2)/(2k^2), in which rows have g.f. (1-2kx+(k^2+1)x^2)/(1-kx)^3.

%H Vincenzo Librandi, <a href="/A081918/b081918.txt">Table of n, a(n) for n = 0..101</a>

%F a(0)=1, a(n)=a(n)=n^n(n^2-n+2n^2)/(2n^2), n>0.

%Y Cf. A000124, A081908, A081909, A081910, A081911, A081912.

%K easy,nonn

%O 0,3

%A _Paul Barry_, Mar 31 2003