The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
It turns out that F541, 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 F2749, 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.
In [1]:
def fibonacci():
a, b = 0, 1
while True:
yield a
a, b = b, a+b
In [2]:
from itertools import islice
list(islice(enumerate(fibonacci()), 10))
Out[2]:
In [3]:
i = 29348710948704982734509283745029348572309458723049
s = str(i)
s0 = s[:9]
s1 = s[-9:]
s0, s1, set(s0), set(s0) - set('0'), len(set(s0) - set('0')), len(set(s0[:5]) - set('0'))
Out[3]:
In [4]:
def is_pandigital(s):
return len(set(s) - set('0')) == 9
In [5]:
def foo():
for i, f in enumerate(fibonacci()):
s = str(f)
if is_pandigital(s[-9:]):
return i
In [6]:
foo()
Out[6]:
In [7]:
def foo():
for i, f in enumerate(fibonacci()):
s = str(f)
if is_pandigital(s[:9]):
return i
In [8]:
foo()
Out[8]:
In [9]:
def foo():
for i, f in enumerate(fibonacci()):
s = str(f)
if is_pandigital(s[:9]) and is_pandigital(s[-9:]):
return i
In [10]:
# foo()
# This took far too long, so I had to refactor for speed.
In [11]:
def foo():
n = 10 ** 9
for i, f in enumerate(fibonacci()):
if is_pandigital(str(f % n)) and is_pandigital(str(f)[:9]):
return i
In [12]:
%timeit foo()
foo()
Out[12]: