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a(1)=0, a(2)=1, a(n+2) = (8*n^2+2*n+1)*a(n+1) - 2*n*(2*n-1)^3*a(n).
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%I #21 Mar 25 2024 12:06:12

%S 0,1,11,299,15371,1285371,159158691,27376820379,6246962876475,

%T 1826295061189275,665694890795056275,296004348848796457275,

%U 157710301268790933578475,99189386694727572925906875

%N a(1)=0, a(2)=1, a(n+2) = (8*n^2+2*n+1)*a(n+1) - 2*n*(2*n-1)^3*a(n).

%H Ofer Yifrach-Stav, <a href="https://hal.science/tel-04513104">Fast and Private Pool Testing and Contributions to Experimental Mathematics</a>, Doctoral thesis, École normale supérieure (Paris, France), HAL Science [math.cs] 2024, Art. No. tel-04513104. See p. 222.

%F a(n+1) = (2*n)!*(2*G*binomial(2*n, n)/4^n - Integral_{t=0..oo} t/cosh(t)^(2*n+1) dt) where G = 0.915965594... is Catalan's constant.

%F a(n) = (2*n-4)! + (2*n-3)^2*a(n-1) for n = 2, 3, ... with a(1) = 0. - _Johannes W. Meijer_, May 24 2009

%t RecurrenceTable[{a[1]==0,a[2]==1,a[n]==(8(n-2)^2+2(n-2)+1)a[n-1]- 2(n-2)(2(n-2)-1)^3 a[n-2]},a,{n,20}] (* _Harvey P. Dale_, May 06 2013 *)

%o (PARI) a(n)=if(n<3,(n+1)%2,(8*(n-2)^2+2*(n-2)+1)*a(n-1)-2*(n-2)*(2*(n-2)-1)^3*a(n-2)) \\ _Benoit Cloitre_, Dec 02 2005

%Y Cf. A006752.

%Y For n >= 1, equals the first left hand column of the Beta triangle A160480. The second left hand column is A160482. - _Johannes W. Meijer_, May 24 2009

%K nonn

%O 1,3

%A _Benoit Cloitre_, Dec 18 2004