A058051 -id:A058051 - cp's OEIS frontend

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

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

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

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

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}.

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

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

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 *)

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

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

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

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

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

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

A274996 a(n) = F(F(F(n))) mod F(F(n)), where F = Fibonacci = A000045.

Original entry on oeis.org

0, 0, 0, 1, 0, 5, 232, 987, 1, 5, 1, 0, 2211236406303914545699412969744873993387956988652, 2211236406303914545699412969744873993387956988653, 139583862445

Offset: 1

Alois P. Heinz, Nov 11 2016

nonn

Alois P. Heinz, Table of n, a(n) for n = 1..19

Cf. A000045, A007570, A058051, A263101, A263112, A338889.

F:= proc(n) local r, M, p; r, M, p:=
      <<1|0>, <0|1>>, <<0|1>, <1|1>>, n;
      do if irem(p, 2, 'p')=1 then r:=
        `if`(nargs=1, r.M, r.M mod args[2]) fi;
         if p=0 then break fi; M:=
        `if`(nargs=1, M.M, M.M mod args[2])
      od; r[1, 2]
    end:
a:= n-> (h-> F(h$2))(F(F(n))):
seq(a(n), n=1..15);

a(n) = A058051(n) mod A007570(n).

A262361 a(n) = L(L(L(n))), where L(n) are Lucas numbers A000032.

Original entry on oeis.org

4, 1, 7, 29, 1149851, 387739824812222466915538827541705412334749

Offset: 0

Alois P. Heinz, Nov 09 2016

nonn

a(6) = 3393011755..4322744978 has 1208 decimal digits and a(7) = 4437405101..8830136999 has 240305 decimal digits.

Alois P. Heinz, Table of n, a(n) for n = 0..5

Cf. A000032, A005371, A058051.

L:= n-> (<<0|1>, <1|1>>^n. <<2, 1>>)[1$2]:
a:= n-> L(L(L(n))):
seq(a(n), n=0..5);

A262361 = Nest[LucasL, #, 3] &; Array[A262361, 6, 0] (* JungHwan Min, Nov 09 2016 *)

from sympy import lucas as L
def a(n):  return L(L(L(n)))
print([a(n) for n in range(6)]) # Michael S. Branicky, Apr 01 2021

A273400 a(n) = Catalan(Catalan(Catalan(n))).

Original entry on oeis.org

1, 1, 2, 39044429911904443959240

Offset: 0

Waldemar Puszkarz, May 21 2016

nonn

a(4) has 1610164 digits and it is thus too large to be included.
Conjecture. The number of digits of a(n) grows asymptotically faster than Catalan(n), i.e., only a finite number of terms of a(n) has the number of digits less than the value of Catalan(n).
This also appears to be true for the Fibonacci sequence (A000045) and the sequence of powers of 2 (A000079): it takes two additional iterations of these sequences for the number of digits of these iterated sequences to grow faster than the corresponding original sequences. However, it appears that it takes only one additional iteration of the factorial (A000142) for this to happen.
The number of digits of a(n) grows asymptotically faster than Fibonacci(n), but that is already true for Catalan(Catalan(n)) (A273399).

			For n = 2, a(2) = Catalan(Catalan(Catalan(2))) = Catalan(Catalan(2)) = Catalan(2) = 2 as Catalan(2) = 2.

Cf. A000108 (Catalan), A273399 (Catalan(Catalan(n))), A058051.

a:= ((n-> binomial(2*n, n)/(n+1))@@3):
seq(a(n), n=0..3);  # Alois P. Heinz, May 27 2025

CatalanNumber[CatalanNumber[CatalanNumber[Range[0, 3]]]]

for(n=0, 3, cn=binomial(2*n, n)/(n+1); cn2=binomial(2*cn, cn)/(cn+1); cn3=binomial(2*cn2, cn2)/(cn2+1); print1(cn3 ", "))

a(n) = A000108(A000108(A000108(n))).

cp's OEIS Frontend

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

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A274996 a(n) = F(F(F(n))) mod F(F(n)), where F = Fibonacci = A000045.

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A262361 a(n) = L(L(L(n))), where L(n) are Lucas numbers A000032.

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A273400 a(n) = Catalan(Catalan(Catalan(n))).

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