Find the index of a number in the fibonacci sequence - bash

Question

I want to find the index of a number in the fibonacci sequence using recursion.
Example : 13 (my input) is the 8th (output) number of the fibonacci sequence
Here is my function :
#!/bin/bash
function fib() {
number=1
cpt=1
if [[ $number -lt $1 ]]
then
compteur=$(($cpt+ 1))
echo $(( $(fib $(($number - 1)) ) + $(fib $(($number - 2)) ) ))
fi
echo $cpt
}
fib $1
I have this error : Wrong symbol near the +
Then I don't know if my logic for the algorithm is good. I don't know if my syntax of how we are calling the function in the recursion in bash is great.
Any help would be appreciated. Thanks !

Answer 1

With recursion, would you please try the following:
#!/bin/bash
# "fib n" returns the nth Fibonacci number (n starts from 0)
fib() {
local n=$1
if (( n == 0 )); then
echo "0"
elif (( n == 1 )); then
echo "1"
else
echo "$(( $(fib "$(( n - 1 ))" ) + $(fib "$(( n - 2 ))" ) ))"
fi
}
input=$1 # user's input as a 1st argument to this script
for (( i = 0; ; i++ )); do # increment the index i starting from 0
f=$(fib i)
(( f >= input)) && break # if "fib i" >= input, then exit the loop
done
if (( f == input )); then # the input is a Fibonacci number
echo "$i"
else
echo "$i (input $input is not a Fibonacci number.)"
fi
By convention, the index starts with "0", then 13 is the 7th Fibonacci number.
Btw if you increase the input number, you'll see the execution of the script takes a while (e.g. please try with 987 as an input).
Here is an alternative without the recursion:
#!/bin/bash
f0=0
f1=1
input=$1
for (( i = 0; ; i++ )); do
(( f0 >= input )) && break
f2=$(( f1 + f0 ))
f0=$f1
f1=$f2
done
if (( f0 == input )); then
echo "$i"
else
echo "$i (input $input is not a Fibonacci number.)"
fi
Although the recursion is mathematically simple and elegant, the number
of computation depends on the implementation. Fibonacci sequence is a
typical example which requires a consideration.
The 1st script includes two problems:
In the 1st recursion, the fib function calls two fib functions.
In the 2nd recursion, four functions are called. The number of computation
reaches O(2**N).
The intermediate values such as fib(n-1) and fib(n-2) are discarded and
calculated again in the next invocation.