A007570 a(n) = F(F(n)), where F is a Fibonacci number.

0, 1, 1, 1, 2, 5, 21, 233, 10946, 5702887, 139583862445, 1779979416004714189, 555565404224292694404015791808, 2211236406303914545699412969744873993387956988653, 2746979206949941983182302875628764119171817307595766156998135811615145905740557

Offset: 0

Author

N. J. A. Sloane, Robert G. Wilson v

Keywords

• nonn
• nice
• easy

Comments

a(20) is approximately 2.830748520089124 * 10^1413, much too large to include even in the b-file. - Alonso del Arte, Apr 30 2020

Let M(0) denote the 2 X 2 identity matrix, and let M(1) = [[0, 1], [1, 1]]. Let M(n) = M(n-2) * M(n-1). Then a(n) is equal to both the (1, 2)-entry and the (2, 1)-entry of M(n). - John M. Campbell, Jul 02 2021

This is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. - Peter Bala, Dec 06 2022

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..19 (terms n = 0..17 from T. D. Noe)
• Alonso del Arte, Table of n, a(n) for n = 0 .. 24, with digits grouped in hundreds
• John M. Campbell, A Matrix-Based Recursion Relation for F_{F_n}, Fib. Quart., Vol. 60, No. 3 (2022), pp. 256-261.
• Bakir Farhi, Summation of certain infinite Fibonacci related series, arXiv:1512.09033 [math.NT], 2015. See p. 6, eq. 2.9.
• George Ledin, Jr., Problem H-147, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 6, No. 6 (1968), p. 352; Converging Fractions, Solution to Problem H-147 by David Zeitlin, ibid., Vol. 8, No. 4 (1970), pp. 387-389.
• Edward A. Parberry, Two recursion relations for F(F(n)), Fib. Quart., Vol. 15, No. 2 (1977), p. 122 and p. 139.
• Martin Stein, Algebraic independence results for reciprocal sums of Fibonacci and Lucas numbers, Dissertation, Hannover: Gottfried Wilhelm Leibniz Universität Hannover, 2012.
• Chris Street, A Recurrence for the Sequence {F(F(n)), n>=0}.

Crossrefs

Cf. A000045, A001622, A005371, A058051, A094214.

Programs

Maple

F:= n-> (<<0|1>, <1|1>>^n)[1, 2]:
a:= n-> F(F(n)):
seq(a(n), n=0..14);  # Alois P. Heinz, Oct 09 2015

Mathematica

F[0] = 0; F[1] = 1; F[n_] := F[n] = F[n - 1] + F[n - 2]; Table[F[F[n]], {n, 0, 14}]
Fibonacci[Fibonacci[Range[0, 20]]] (* Harvey P. Dale, May 05 2012 *)

PARI

a(n)=fibonacci(fibonacci(n)) \\ Charles R Greathouse IV, Feb 03 2014

Python

from sympy import fibonacci
def a(n): return fibonacci(fibonacci(n))
print([a(n) for n in range(15)]) # Michael S. Branicky, Feb 02 2022

Sage

[fibonacci(fibonacci(n)) for n in range(0, 14)] # Zerinvary Lajos, Nov 30 2009

Scala

val fibo: LazyList[BigInt] = (0: BigInt) #:: (1: BigInt) #:: fibo.zip(fibo.tail).map { n => n._1 + n._2 }
val fiboLimited: LazyList[Int] = 0 #:: 1 #:: fiboLimited.zip(fiboLimited.tail).map { n => n._1 + n._2 } // Limited to 32-bit integers because that's the type for LazyList apply()
(0 to 19).map(n => fibo(fiboLimited(n))) // Alonso del Arte, Apr 30 2020

Formula

a(n+1)/a(n) ~ phi^(F(n-1)), with phi = (1 + sqrt(5))/2 = A001622. - Carmine Suriano, Jan 24 2011

Sum_{n>=1} 1/a(n) = 3.7520024260... is transcendental (Stein, 2012). - Amiram Eldar, Oct 30 2020

Sum_{n>=1} (-1)^(F(n)+1)*a(n-1)/(a(n)*a(n+1)) = 1/phi (A094214) (Farhi, 2015). - Amiram Eldar, Apr 07 2021

Limit_{n->oo} a(n+1)/a(n)^phi = 5^((phi-1)/2) = 1.6443475285..., where phi is the golden ratio (A001622) (Ledin, 1968) - Amiram Eldar, Feb 02 2022

Extensions

One more term from Harvey P. Dale, May 05 2012