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A305459 a(0) = 1, a(1) = 3, a(n) = 3*n*a(n-1) + a(n-2).

%I A305459 #22 Aug 27 2019 17:45:15
%S A305459 1,3,19,174,2107,31779,574129,12088488,290697841,7860930195,
%T A305459 236118603691,7799774851998,281028013275619,10967892292601139,
%U A305459 460932504302523457,20752930585906156704,996601600627798045249,50847434562603606464403
%N A305459 a(0) = 1, a(1) = 3, a(n) = 3*n*a(n-1) + a(n-2).
%C A305459 Let S(i,j,n) denote a sequence of the form a(0) = 1, a(1) = i, a(n) = i*n*a(n-1) + j*a(n-2). Then S(i,j,n) = Sum_{k=0..floor(n/2)} ((n-k)!/k!)*binomial(n-k,k)*i^(n-2*k)*j^k.
%H A305459 Seiichi Manyama, <a href="/A305459/b305459.txt">Table of n, a(n) for n = 0..380</a>
%F A305459 a(n) = Sum_{k=0..floor(n/2)} ((n-k)!/k!)*binomial(n-k,k)*3^(n-2*k).
%F A305459 a(n) ~ BesselI(0, 2/3) * n! * 3^n. - _Vaclav Kotesovec_, Jun 03 2018
%p A305459 a:=proc(n) option remember: if n=0 then 1 elif n=1 then 3 elif n>=2 then 3*n*procname(n-1)-procname(n-2) fi; end:
%p A305459 seq(a(n),n=0..20); # _Muniru A Asiru_, Jun 01 2018
%t A305459 RecurrenceTable[{a[0]==1,a[1]==3,a[n]==3n a[n-1]+a[n-2]},a,{n,20}] (* _Harvey P. Dale_, Aug 27 2019 *)
%o A305459 (PARI) {a(n) = sum(k=0, n/2, ((n-k)!/k!)*binomial(n-k,k)*3^(n-2*k))}
%o A305459 (GAP) List([0..20],n->Sum([0..Int(n/2)],k->((Factorial(n-k))/(Factorial(k))*Binomial(n-k,k)*3^(n-2*k)))); # _Muniru A Asiru_, Jun 01 2018
%Y A305459 Cf. A001040, A036243, A213190, A305460.
%K A305459 nonn
%O A305459 0,2
%A A305459 _Seiichi Manyama_, Jun 01 2018