@Razetime I was struggling to find a good way to solve this, but then I saw your deleted solution to the lucky factors challenge, and it seems great! Some fixes are needed to make it properly compute the lucky factors, but you should definitely post it to the lucky numbers challenge

@Leo yeah, that what I was trying with ¡o§,o←!2ΣFomhC,2N

Not sure I can understand that, but this solves the lucky numbers challenge nicely

i think it might fail on higher tests

the main thing is I'm making a tuple (step,seq)

each time I update the step by taking the first number of the second slice, and then mhC on the sequence.

@Razetime Got it now, but I think this is not right

The step should be the first number still in the sequence that is bigger than the previous step

idk, seems to fit the description in caird's challange

@Leo exactly, which is the first number in the second slice

@Razetime I think your steps would go [2,3,7,21...] rather than [2,3,7,9..]

hm.

well I can't really check it since it never seems to infer so lol

@Razetime Confirmed

so how does the folding startegy work?

@Razetime Why do you ask me, you wrote it :P

well you changed it so I'm not too sure of the logic

(jk)

I think the idea in there is that except for 2, the steps are exactly the numbers in the final sequence

So the 2 is managed separately by starting directly from the odd numbers, then the fold goes through the indices from 2 onward and uses the nth element of the sequence as the step each time

huh so my approach was mistakenly right

nice

Mistakenly right is the second best kind of right

overall big win