Yes, quite right! You're my conscience as ever, Raul.
Sorry - I misremembered and didn't check.
Moebius mu(n) is 1 for 3 and 5, _1 for 3*5=15, and 0 for powers higher
than one of 3 and 5 and their products, so (eg) 0 for 25, 27 etc.
Also, 99 is the upper inclusive bound.
So:
+/99 (]*2!>:@(<.@%|)) 3 5 _15 NB. correctly hand-coded moebius!
2318
cf
+/I.+./3 5 (0=|)"0/i.100 NB. Direct method
2318
Right idea. Previously almost completely wrong example!!!
I hope it hasn't put Geoff off.
Mike
On 05/05/2016 14:31, Raul Miller wrote:
On Thu, May 5, 2016 at 4:31 AM, Mike Day <mike_liz....@tiscali.co.uk> wrote:
+/1683 1050 _594 315 _250 162 _135
2231
Er... is that the right answer, somehow?
Thanks,
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