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

%I #12 Sep 26 2021 03:19:04

%S 1,2,4,14,54,292,1696,12500,97300,909800,8888400,100778200,1185163000,

%T 15798002000,217275712000,3329482106000,52433793018000,

%U 908110694260000,16113910669480000,311249886303980000,6144485548655740000

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

%D Studied by M. Bhargava (bhargava(AT)math.Princeton.EDU) and H. S. Wilf.

%H T. D. Noe, <a href="/A047990/b047990.txt">Table of n, a(n) for n=0..100</a>

%F a(n) ~ c * n!, where c = sqrt(Pi) * (1/(Gamma(1/2 + i/2) * Gamma(1 - i/2)) + 1/(Gamma(1/2 - i/2) * Gamma(1 + i/2)))/2 + hypergeom([1 - i, 1 - i], [2], -1) / 2^i = 2.6068398732089369139523868268495087173767147472699250704975984714... and i is the imaginary unit. - _Vaclav Kotesovec_, Sep 26 2021

%t RecurrenceTable[{a[0]==1,a[1]==2,a[n]==a[n-1]+((n-1)^2+1)a[n-2]},a,{n,20}] (* _Harvey P. Dale_, May 07 2013 *)

%Y Cf. A047991.

%K nonn,easy,nice

%O 0,2

%A _N. J. A. Sloane_