The Fibonacci sequence is defined by the recurrence relation:

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

Hence the first 12 terms will be: $$ F_1 = 1\\ F_2 = 1\\ F_3 = 2\\ F_4 = 3\\ F_5 = 5\\ F_6 = 8\\ F_7 = 13\\ F_8 = 21\\ F_9 = 34\\ F_{10} = 55\\ F_{11} = 89\\ F_{12} = 144\\ $$ The 12th term, F12, is the first term to contain three digits.

What is the first term in the Fibonacci sequence to contain 1000 digits?



In [12]:

    
def fib():
    a = 0
    b = 1
    yield b
    while True:
        a, b = b, a + b
        yield b



In [13]:

    
n = 12
for at, f in enumerate(fib(), start=1):
    print(at, f)
    if at == n:
        break



In [15]:

    
def at_least_n(n):
    for at, f in enumerate(fib(), start=1):
        l = len(str(f))
        if l >= n:
            return at, l , f
            break



In [16]:

    
at_least_n(3)









    Out[16]:





(12, 3, 144)



In [17]:

    
at_least_n(1000)









    Out[17]:





(4782,
 1000,
 1070066266382758936764980584457396885083683896632151665013235203375314520604694040621889147582489792657804694888177591957484336466672569959512996030461262748092482186144069433051234774442750273781753087579391666192149259186759553966422837148943113074699503439547001985432609723067290192870526447243726117715821825548491120525013201478612965931381792235559657452039506137551467837543229119602129934048260706175397706847068202895486902666185435124521900369480641357447470911707619766945691070098024393439617474103736912503231365532164773697023167755051595173518460579954919410967778373229665796581646513903488154256310184224190259846088000110186255550245493937113651657039447629584714548523425950428582425306083544435428212611008992863795048006894330309773217834864543113205765659868456288616808718693835297350643986297640660000723562917905207051164077614812491885830945940566688339109350944456576357666151619317753792891661581327159616877487983821820492520348473874384736771934512787029218636250627816)