Is there any way to know the nth digit of k! without having to calculate the factorial?
And, for some k, when k! is prime? (disregarding the trailing zeros, and k>2)
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1$k!$ is never prime since $2$ is always a divisor. $2!$ is the only exception. – Chubby Chef Nov 26 '20 at 22:45
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for all $k \ge 3$, $k!$ is divisible by $3$ (even after getting rid of the trailing zero). – achille hui Nov 26 '20 at 22:48
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Chubby, but is to ignore the trailing zeros, doesn't change? – thiago Nov 26 '20 at 22:54
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achille, got it – thiago Nov 26 '20 at 22:54
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but the first question, do you know something about it? – thiago Nov 26 '20 at 22:55
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See https://oeis.org/A008905 for the leading digit. (Curiously, the sequence $4,2,2,0,0,6,6,9\ldots$ of second digits is not -- yet -- in the OEIS.) – Barry Cipra Nov 26 '20 at 23:10
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hmmm interesting – thiago Nov 26 '20 at 23:17
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are there sequences for third, fourth, ... digit? – thiago Nov 26 '20 at 23:18
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@thiago Doesn't matter if you take the trailing zeroes into consideration or not. This has to do with the fundamental theorem of arithmetic. – Chubby Chef Nov 26 '20 at 23:43