7 ms·
You gave it away with a slight error - the sequence is 0, 1, 2, 720... For the curious, the sequence follows factorial (!) applied twice, i.e. (0!)!, (1!)! etc
by civild 14y ago
You gave it away with a slight error - the sequence is 0, 1, 2, 720...
For the curious, the sequence follows factorial (!) applied twice, i.e. (0!)!, (1!)! etc. (3!)! is 720, and (4!)! is 6.20448402 × 10^23.
- anonymoushn 14y agoI have (0!)! = (1!)! = 1 over here.
- benjoffe 14y agoIn that case the sequence should be 1, 1, 2, 720... as zero factorial equals one. https://www.google.com/search?q=0 https://www.google.com/search?q=0!
- civild 14y agoOops, quite right!
- dedward 14y ago0 is correct pragmatacally, and makes the puzzle harder for those not wise that 0!=1 by the definition of factorial. (as opposed to proof.... factorial being a shorthand for math and ths being convenient. if there were a proof it would be a theorem, which its not.) i knew factorial but i had to read up on 0!, news to me too. on another note, without context, we could say there is an infinite set of functions that satisfy any such question. "what could come next and why" or something. edit: 1 is indeed correct for the sequence, ignore that part... my bad.
- deleted 14y ago[deleted]
- mikeash 14y agoHis sequence (corrected to 1, 1, 2, 720!, ...) would work for the factorial applied three times. Although the ! pretty much gives it away then.
- reitzensteinm 14y agoWhile the zero is off in the GP's post, he technically could have been talking about the series of n!!!
- waqf 14y agoNo, the sequence is correct as originally posted and the rule is: 0, 1!, (2!)!, ((3!)!)!, (((4!)!)!)!, ...