How many prime factors does the 1801st Fibonacci number have?
4
110
Ṁ103Ṁ180
2100
1D
1W
1M
ALL
34%
2
18%
3
15%
4
34%
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https://r-knott.surrey.ac.uk/Fibonacci/fibtable.html almost all of them have more than 3
@nanob0nus Indeed, the prime number theorem tells us that the density of primes goes like 1/log(n), and knth fibonacci number is divisible by the nth, so for any x, almost all Fibonacci numbers will have at least x factors.
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