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%I #18 Nov 11 2018 20:40:01
%S 0,1,1,2,9,1241,2298661010,19127219051011953293860241761,
%T 8789440239853164630485833302292601093162389737995133605845884014903267053091248194081
%N a(n) = a(n-1)^3 + a(n-2)^9, a(0) = 0, a(1) = 1.
%C In general, if a(n) = a(n-1)^k + a(n-2)^(k^2), k > 1, then a(n) ~ d * c^(k^n), where c is a constant (dependent only on k, a(0) and a(1)), d is the root of the equation d^(k-1) * (1 + d^(k*(k-1))) = 1.
%C From _Massimo Galasi_, Jul 26 2018: (Start)
%C The recursion a(n)=a(n-1)^k+a(n-2)^(k^2) with a(0)=1, and a(1)=2, gives the number of independent sets of the k-ary tree with n levels. (For n=0 one has the empty tree.)
%C Thus this sequence, starting with 1, 2, 9, 1241, ... is the case k=3. Sequence A076725 is the case k=2. For k=1 the tree becomes a path and a(n) is Fibonacci(n+1). (End)
%F a(n) ~ d * c^(3^n), where c = 1.03028886637346769106..., d = 0.85117093406701547... is the root of the equation d^2 + d^8 = 1.
%p a:=proc(n) option remember; if n=0 then 0 else if n=1 then 1 else a(n-1)^3+a(n-2)^9 fi fi end: seq(a(n), n = 0..9);
%t RecurrenceTable[{a[0]==0, a[1]==1, a[n] == a[n-1]^3 + a[n-2]^9}, a, {n, 0, 10}]
%Y Cf. A000278, A076725.
%K nonn
%O 0,4
%A _Vaclav Kotesovec_, Dec 18 2014
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