A143077 - OEIS

%I #21 Apr 26 2020 15:19:55

%S 1,1,4,31,485,27343,3595117,1359551201,1310562076858,3378072688461451,

%T 22702751567715567129,401359405793550977993221,

%U 18572242457139030215454649193,2252593125544789695036793639095505

%N a(n) is the n-th term of a pseudo-Fibonacci sequence created by applying the function fib(1,...,n) to itself n times.

%C This sequence grows faster than any exponential sequence. The implementation here is quite slow.

%C Let g(x) be fib(1,1,x), g returns y; let h(y) be fib(1,y,x), h returns z; let i(z) be z be applied to itself x-1 times. Then f(x) = i(h(g(x))).

%e n=1: fib(1,1,n);

%e n=2: fib(1,fib(1,1,n),n);

%e n=3: fib(1,fib(1,fib(1,1,n),n),n) ...

%e   f(3) is fib(1,fib(1,fib(1,1,3),3),3);

%e   f(3) simplifies to fib(1,fib(1,2,3),3);

%e   f(3) simplifies to fib(1,3,3);

%e   f(3) is 4.

%o (Python)

%o def fib(arb1, arb2, nth):

%o     if nth == 0:

%o         return arb1

%o     if nth == 1:

%o         return arb2

%o     x = [0]*nth

%o     x[0] = arb1

%o     x[1] = arb2

%o     for i in range(2, nth, 1):

%o         x[i] = x[i-1]+x[i-2]

%o     return x[-1]

%o def fib2d(n):

%o     return fib(1, fib(1, 1, n), n)

%o def fib3d(n):

%o     return fib(1, fib(1, fib(1, 1, n), n), n)

%o def slowfibnd(n): # This is an inelegant way to generate a(n)

%o     begin = "fib(1, 0+1, n)"

%o     for x in range(n-1):

%o         begin = begin.replace('0+1', 'fib(1, 0+1, n)')

%o     return eval(begin)

%Y Cf. A000045 is the Fibonacci function fib(1, 1, n), A142975 is the Fibonacci function applied to itself fib(1, fib(1, 1, n), n).

%K nonn

%O 1,3

%A Gregory Nisbet (gregory.nisbet(AT)gmail.com), Jul 22 2008