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a(0)=a(1)=1, a(n)=a(n-2)+(n+1)*a(n-1).
7

%I #35 Jul 26 2024 11:39:11

%S 1,1,4,17,89,551,3946,32119,293017,2962289,32878196,397500641,

%T 5200386529,73202912047,1103244067234,17725107987791,302430079859681,

%U 5461466545462049,104070294443638612,2086867355418234289,43928284758226558681,968509132036402525271

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

%H Vincenzo Librandi, <a href="/A058279/b058279.txt">Table of n, a(n) for n = 0..200</a>

%H Chris Cannings, <a href="http://dx.doi.org/10.4236/am.2013.45105">The Stationary Distributions of a Class of Markov Chains</a>, Applied Mathematics, Vol. 4 No. 5, 2013, pp. 769-773.

%H David Ji, Michael Li, and Daniel Wang, <a href="https://arxiv.org/abs/2407.15889">Parallel chip-firing games on directed graphs</a>, arXiv:2407.15889 [math.CO], 2024.

%F a(n) is asymptotic to c*n! with c=0.9007... - _Benoit Cloitre_, Sep 03 2002

%F Right asymptotic (with offset=0) is a(n) ~ c * (n+1)!, where c = 2*BesselI(1,2)-BesselI(0,2) = 0.9016884069385908593273044... - _Vaclav Kotesovec_, Jan 05 2013

%F E.g.f.: 2*Pi*(I*BesselY(3, 2*I)*BesselI(2, 2*sqrt(1-x)) + BesselI(3, 2)*BesselY(2, 2*I*sqrt(1-x)))/(1-x). Such e.g.f. computations were inspired after e-mail exchange with Gary Detlefs. After differentiation and putting x=0 one has to use simplifications. See the Abramowitz-Stegun handbook, p.360, 9.1.16. [_Wolfdieter Lang_, May 19 2010]

%p A058279 := proc(n) option remember; if n <= 1 then 1 else A058279(n-2)+(n+1)*A058279(n-1); fi; end;

%t RecurrenceTable[{a[0] == a[1] == 1, a[n] == a[n-2] + a[n-1] (n+1)}, a, {n, 30}] (* _Vincenzo Librandi_, May 06 2013 *)

%o (Magma) [n le 2 select 1 else Self(n-2)+Self(n-1)*(n): n in [1..30]]; // _Vincenzo Librandi_, May 06 2013

%Y See A058307 for the same recurrence with 0,1 inputs. [_Wolfdieter Lang_, May 19 2010]

%K nonn

%O 0,3

%A _N. J. A. Sloane_, Dec 09 2000