Pandigital Fibonacci ends

The Fibonacci sequence is defined by the recurrence relation:

F_n = F_n−1 + F_n−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]:





[(0, 0),
 (1, 1),
 (2, 1),
 (3, 2),
 (4, 3),
 (5, 5),
 (6, 8),
 (7, 13),
 (8, 21),
 (9, 34)]



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]:





('293487109',
 '458723049',
 {'0', '1', '2', '3', '4', '7', '8', '9'},
 {'1', '2', '3', '4', '7', '8', '9'},
 7,
 5)



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]:





541



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]:





2749



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()









    



1 loop, best of 3: 32.3 s per loop






    Out[12]:





329468