A252584 a(n) = a(n-1)^3 + a(n-2)^9, a(0) = 0, a(1) = 1.

0, 1, 1, 2, 9, 1241, 2298661010, 19127219051011953293860241761, 8789440239853164630485833302292601093162389737995133605845884014903267053091248194081

Offset: 0

Author

Vaclav Kotesovec, Dec 18 2014

Keywords

• nonn

Comments

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.

From Massimo Galasi, Jul 26 2018: (Start)

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.)

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)

Crossrefs

Cf. A000278, A076725.

Programs

Maple

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);

Mathematica

RecurrenceTable[{a[0]==0, a[1]==1, a[n] == a[n-1]^3 + a[n-2]^9}, a, {n, 0, 10}]

Formula

a(n) ~ d * c^(3^n), where c = 1.03028886637346769106..., d = 0.85117093406701547... is the root of the equation d^2 + d^8 = 1.