A143077 a(n) is the n-th term of a pseudo-Fibonacci sequence created by applying the function fib(1,...,n) to itself n times.

1, 1, 4, 31, 485, 27343, 3595117, 1359551201, 1310562076858, 3378072688461451, 22702751567715567129, 401359405793550977993221, 18572242457139030215454649193, 2252593125544789695036793639095505

Offset

1,3

Comments

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

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))).

Links

Table of n, a(n) for n=1..14.

Example

n=1: fib(1,1,n);
n=2: fib(1,fib(1,1,n),n);
n=3: fib(1,fib(1,fib(1,1,n),n),n) ...
  f(3) is fib(1,fib(1,fib(1,1,3),3),3);
  f(3) simplifies to fib(1,fib(1,2,3),3);
  f(3) simplifies to fib(1,3,3);
  f(3) is 4.

Prog

(Python)
def fib(arb1, arb2, nth):
    if nth == 0:
        return arb1
    if nth == 1:
        return arb2
    x = [0]*nth
    x[0] = arb1
    x[1] = arb2
    for i in range(2, nth, 1):
        x[i] = x[i-1]+x[i-2]
    return x[-1]
def fib2d(n):
    return fib(1, fib(1, 1, n), n)
def fib3d(n):
    return fib(1, fib(1, fib(1, 1, n), n), n)
def slowfibnd(n): # This is an inelegant way to generate a(n)
    begin = "fib(1, 0+1, n)"
    for x in range(n-1):
        begin = begin.replace('0+1', 'fib(1, 0+1, n)')
    return eval(begin)

Crossrefs

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).

Sequence in context: A141827 A262529 A350608 * A203011 A228467 A005841

Adjacent sequences: A143074 A143075 A143076 * A143078 A143079 A143080

Keyword

nonn

Author

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

Status

approved