## The Fibonacci sequence is defined by the recurrence relation:

$F_n = F_{n – 1} + F_{n – 2}$, where $F_1 = 1$ and $F_2 = 1$.

It turns out that $F_{541}$, which contains $113$ digits, is the first Fibonacci number for which the last nine digits are $1$-$9$ pandigital (contain all the digits $1$ to $9$, but not necessarily in order). And $F_{2749}$, which contains $575$ digits, is the first Fibonacci number for which the first nine digits are $1$-$9$ pandigital.

Given that $F_k$ is the first Fibonacci number for which the first nine digits AND the last nine digits are $1$-$9$ pandigital, find $k$.

### To solve this problem, we can generate the Fibonacci sequence iteratively, checking for the given pandigital conditions as we go. We’ll start by defining a function to check if a number is pandigital.

“`python

def is_pandigital(n):

digits = set(str(n))

return len(digits) == 9 and ‘0’ not in digits

“`

Now, we’ll create a loop to generate the Fibonacci sequence until we find a number that satisfies both conditions:

“`python

F1 = F2 = 1

k = 2

while True:

Fn = F1 + F2

k += 1

if is_pandigital(Fn % 1000000000) and is_pandigital(int(str(Fn)[:9])):

break

F1, F2 = F2, Fn

“`

Finally, we print the value of `k`, which represents the index of the first Fibonacci number that satisfies the given pandigital conditions:

“`python

print(k)

“`

Putting it all together, the Python code to solve this problem is:

“`python

def is_pandigital(n):

digits = set(str(n))

return len(digits) == 9 and ‘0’ not in digits

F1 = F2 = 1

k = 2

while True:

Fn = F1 + F2

k += 1

if is_pandigital(Fn % 1000000000) and is_pandigital(int(str(Fn)[:9])):

break

F1, F2 = F2, Fn

print(k)

“`

When you run this code, it will output the value of `k`, which represents the index of the first Fibonacci number where both the last nine digits and the first nine digits are pandigital.

##### More Answers:

Optimum PolynomialTriangle Containment

Special Subset Sums: Optimum